NCERT Solutions for Class 11 Maths Chapter 13 - In Hindi

प्रश्नावली 13.1 

निम्नलिखित सीमाओ के मान प्राप्त कीजिए 

1. \[\mathop {\lim }\limits_{{\text{x}} \to 3} {\text{x + 3}}\]

उत्तर: \[\mathop {\lim }\limits_{{\text{x}} \to 3} {\text{(x + 3)}}\;{\text{ = }}\;{\text{3 + 3}}\;{\text{ = }}\;{\text{6}}\]

2. \[\mathop {\lim }\limits_{{\text{x}} \to 1} {\text{(x - }}\dfrac{{22}}{7})\]

उत्तर: \[\mathop {\lim }\limits_{{\text{x}} \to 1} {\text{(x - }}\dfrac{{22}}{7})\; = \;\mathop {\lim }\limits_{{\text{x}} \to \pi } {\text{(x - }}\dfrac{{22}}{7})\; = \;\pi {\text{ - }}\dfrac{{22}}{7}\]

3. \[\mathop {\lim }\limits_{{\text{r}} \to 1} {{(\pi }}{{\text{r}}^{\text{2}}})\]

उत्तर:\[\mathop {\lim }\limits_{{\text{r}} \to 1} {{(\pi }}{{\text{r}}^{\text{2}}})\] $= \pi {(1)^2} = \pi $

4. \[\mathop {\lim }\limits_{{\text{x}} \to 4} {\text{(}}\dfrac{{{\text{4x + 3}}}}{{{\text{x - 2}}}})\]

उत्तर: \[\mathop {\lim }\limits_{{\text{x}} \to 4} {\text{(}}\dfrac{{{\text{4x + 3}}}}{{{\text{x - 2}}}})\; = \;\dfrac{{{\text{4(4) + 3}}}}{{{\text{(4) - 2}}}}\; = \;\dfrac{{16 + 3}}{2}\; = \;\dfrac{{19}}{2}\]

5. \[\mathop {\lim }\limits_{{\text{x}} \to  - 1} {\text{(}}\dfrac{{{{\text{x}}^{{\text{10}}}}{\text{ + }}{{\text{x}}^{\text{5}}}{\text{ + 1}}}}{{{\text{x - 1}}}})\]

उत्तर: \[\mathop {\lim }\limits_{{\text{x}} \to  - 1} {\text{(}}\dfrac{{{{\text{x}}^{{\text{10}}}}{\text{ + }}{{\text{x}}^{\text{5}}}{\text{ + 1}}}}{{{\text{x - 1}}}})\; = \;\dfrac{{{{( - 1)}^{{\text{10}}}}{\text{ + ( - 1}}{{\text{)}}^{\text{5}}}{\text{ + 1}}}}{{{\text{( - 1) - 1}}}}\]

\[\dfrac{{1 - 1 + 1}}{{ - 2}}\; = \;\dfrac{{ - 1}}{2}\]

6. \[\mathop {\lim }\limits_{{\text{x}} \to 0} {\text{(}}\dfrac{{{{{\text{(x + 1)}}}^5}{\text{ - 1}}}}{{\text{x}}})\]

उत्तर: \[\mathop {\lim }\limits_{{\text{x}} \to 0} {\text{(}}\dfrac{{{\text{(1 + 5x + 10}}{{\text{x}}^2}{\text{ + 10}}{{\text{x}}^3}{\text{ + 5}}{{\text{x}}^4}{\text{ + }}{{\text{x}}^5}{\text{) - 1}}}}{{\text{x}}})\; = \;\mathop {\lim }\limits_{{\text{x}} \to 0} {\text{(}}\dfrac{{{\text{x(5 + 10x + 10}}{{\text{x}}^2}{\text{ + 5}}{{\text{x}}^3}{\text{ + }}{{\text{x}}^4}{\text{)}}}}{{\text{x}}})\]

= \[\mathop {\lim }\limits_{{\text{x}} \to 0} {\text{(5 + 10x + 10}}{{\text{x}}^{\text{2}}}{\text{ + 5}}{{\text{x}}^{\text{3}}}{\text{ + }}{{\text{x}}^{\text{4}}}{\text{)}}\]

=5

या

हम जानते है 

$  \mathop {\lim }\limits_{{\text{x}} \to a} \dfrac{{{{\text{x}}^{\text{n}}}{\text{ - 1}}}}{{{\text{x - 1}}}}\;{\text{ = }}\;{\text{n}}{{\text{d}}^{{\text{n - 1}}}}\; $

$   = \;\mathop {\lim }\limits_{{\text{x}} \to 0} \dfrac{{{{{\text{(x + 1)}}}^{\text{5}}}{\text{ - 1}}}}{{\text{x}}}\; $

$   = \;\mathop {\lim }\limits_{{\text{x}} \to 0} \dfrac{{{{{\text{(x + 1)}}}^5} - 1}}{{({\text{x + 1}}) - 1}}\; $

$   = \;\mathop {\lim }\limits_{{\text{x}} \to 0} 5{({\text{x}} + 1)^4}\; $

$   = \;5 $

7. \[\mathop {\lim }\limits_{{\text{x}} \to 2} {\text{(}}\dfrac{{{\text{3}}{{\text{x}}^2}{\text{ - x - 10}}}}{{{{\text{x}}^2} - 4}})\]

उत्तर: \[{\text{x}}\;{\text{ = }}\;{\text{2}}\] पर, दिए गए तर्कसंगत फंगक्शन का मान रूप लेता है। 

$  = \;\mathop {\lim }\limits_{{\text{x}} \to 2} \dfrac{{{\text{3}}{{\text{x}}^{\text{2}}}{\text{ - x - 10}}}}{{{{\text{x}}^{\text{2}}}{\text{ - 4}}}} $

$   = \;\mathop {\lim }\limits_{{\text{x}} \to 2} \dfrac{{{\text{(x - 2)(3x + 5)}}}}{{{\text{(x - 2)(x + 2)}}}} $

$   = \;\mathop {\lim }\limits_{{\text{x}} \to 2} \dfrac{{{\text{3x + 5}}}}{{{\text{x + 2}}}} $

$   = \;\dfrac{{3(2) + 5}}{{2 + 2}}\; = \;\dfrac{{11}}{4} $

8. \[\mathop {\lim }\limits_{{\text{x}} \to 3} {\text{(}}\dfrac{{{{\text{x}}^4}{\text{ - 81}}}}{{{\text{2}}{{\text{x}}^{\text{2}}}{\text{ - 5x - 3}}}})\]

उत्तर: \[\mathop {\lim }\limits_{{\text{x}} \to 3} \dfrac{{{\text{(x - 3)(x + 3)}}\left( {{{\text{x}}^{\text{2}}}{\text{ + 9}}} \right)}}{{{\text{(x - 3)(2x + 1)}}}}\]

$   = \;\dfrac{{(3 + 3)\left( {{3^2} + 9} \right)}}{{2(3) + 1}} $

$   = \;\dfrac{{(3 + 3)(9 + 9)}}{{(6 + 1)}} $

$   = \;\dfrac{{6 \times 18}}{7} $

$   = \;\dfrac{{108}}{7} $

9. \[\mathop {\lim }\limits_{{\text{x}} \to 0} {\text{(}}\dfrac{{{\text{ax + b}}}}{{{\text{cx + 1}}}})\]

उत्तर: \[\dfrac{{{\text{a(0) + b}}}}{{{\text{c(0) + 1}}}}\; = \;\dfrac{{{\text{0 + b}}}}{{{\text{0 + 1}}}}\; = \;{\text{b}}\]

10. \[\mathop {\lim }\limits_{{\text{z}} \to 1} {\text{(}}\dfrac{{{{\text{z}}^{\dfrac{{\text{1}}}{{\text{3}}}}}{\text{ - 1}}}}{{{{\text{z}}^{\dfrac{{\text{1}}}{6}}}{\text{ - 1}}}})\]

उत्तर: \[\mathop {\lim }\limits_{{\text{z}} \to 1} \dfrac{{{\text{(}}{{\text{z}}^{\dfrac{{\text{1}}}{6}}}{\text{ - 1)(}}{{\text{z}}^{\dfrac{{\text{1}}}{6}}}{\text{ + 1)}}}}{{{{\text{z}}^6}{\text{ - 1}}}}\]

= \[\dfrac{{1 + 1}}{1}\; = \;2\]

11. \[\mathop {\lim }\limits_{{\text{x}} \to 1} {\text{(}}\dfrac{{{\text{a}}{{\text{x}}^{\text{2}}}{\text{ + bx + c}}}}{{{\text{c}}{{\text{x}}^{\text{2}}}{\text{ + bx + a}}}}),\;{\text{a + b + c}}\; \ne \;0\]

उत्तर: \[\dfrac{{{\text{a(1}}{{\text{)}}^{\text{2}}}{\text{ + b(1) + c}}}}{{{\text{c(1}}{{\text{)}}^{\text{2}}}{\text{ + b(1) + a}}}}\]

= \[\dfrac{{{\text{a + b + c}}}}{{{\text{c + b + a}}}}\;{\text{ = }}\;{\text{1}}\]

12. \[\mathop {\lim }\limits_{{\text{x}} \to  - 2} {\text{(}}\dfrac{{\dfrac{{\text{1}}}{{\text{x}}}{\text{ + }}\dfrac{{\text{1}}}{{\text{2}}}}}{{{\text{x + 2}}}}{\text{)}}\]

उत्तर: \[\mathop {\lim }\limits_{{\text{x}} \to  - 2} \dfrac{{{\text{x + 2}}}}{{{\text{2x(x + 2)}}}}\]

\[\mathop {\lim }\limits_{{\text{x}} \to  - 2} \dfrac{{\text{1}}}{{{\text{2x}}}}\;{\text{ = }}\;\mathop {\lim }\limits_{{\text{x}} \to  - 2} \dfrac{{\text{1}}}{{{\text{2( - 2)}}}}\; = \; - \dfrac{1}{4}\]

13. \[\mathop {\lim }\limits_{{\text{x}} \to 0} {\text{(}}\dfrac{{{\text{sinax}}}}{{{\text{bx}}}}{\text{)}}\]

उत्तर: \[\mathop {\lim }\limits_{{\text{x}} \to 0} \dfrac{{{\text{sinax}}}}{{{\text{ax}}}}.\dfrac{{\text{a}}}{{\text{b}}}\]     ( \[\because \;\mathop {\lim }\limits_{{\text{x}} \to 0} \dfrac{{{\text{sinax}}}}{{{\text{ax}}}}\; = \;1\] )

= \[1 \times \dfrac{{\text{a}}}{{\text{b}}}\; = \;\dfrac{{\text{a}}}{{\text{b}}}\]

14. \[\mathop {\lim }\limits_{{\text{x}} \to 0} {\text{(}}\dfrac{{{\text{sinax}}}}{{{\text{sinbx}}}}{\text{),}}\;{\text{a,b}}\; \ne \;{\text{0}}\]

उत्तर: \[\mathop {\lim }\limits_{{\text{x}} \to 0} {\text{(}}\dfrac{{{\text{sinax}}}}{{{\text{ax}}}}{\text{)(}}\dfrac{{{\text{bx}}}}{{{\text{sinbx}}}}{\text{)}} \times \dfrac{{\text{a}}}{{\text{b}}}\]

\[1 \times 1 \times \dfrac{{\text{a}}}{{\text{b}}}\; = \;\dfrac{{\text{a}}}{{\text{b}}}\]

15. \[\mathop {\lim }\limits_{{\text{x}} \to {{\pi }}} {\text{(}}\dfrac{{{{sin(\pi  - x)}}}}{{{{\pi (\pi  - x)}}}}{\text{)}}\]

उत्तर: \[\pi  - {\text{x}}\; = \;\theta \;(\therefore \;{\text{x}} \to \pi ,\theta  \to 0)\]

$   = \;\mathop {\lim }\limits_{{\text{x}} \to \pi } \dfrac{1}{\pi }\dfrac{{{{sin(\pi  - x)}}}}{{{{(\pi  - x)}}}} $

$   = \;\mathop {\lim }\limits_{\theta  \to 0} \dfrac{{\sin \theta }}{{\pi \theta}}$ 

$   = \;\mathop {\lim }\limits_{\theta  \to 0} \left( {\dfrac{{\sin \theta }}{{\pi \theta }}} \right)\dfrac{1}{\pi }\; = \;\dfrac{1}{\pi } $

16. \[\mathop {\lim }\limits_{{\text{x}} \to 0} {\text{(}}\dfrac{{{\text{cosx}}}}{{{{\pi  - x}}}}{\text{)}}\]

उत्तर: \[\dfrac{{{\text{cos0 }}}}{{{{\pi  - 0 }}}}\; = \;\dfrac{1}{\pi }\]

17. \[\mathop {\lim }\limits_{{\text{x}} \to 0} \dfrac{{{\text{cos2x - 1}}}}{{{\text{cosx - 1}}}}\]

उत्तर: \[\mathop {\lim }\limits_{{\text{x}} \to 0} \dfrac{{{\text{1 - 2si}}{{\text{n}}^2}{\text{x - 1}}}}{{{\text{cosx - 1}}}}\]

$  = \;\mathop {\lim }\limits_{{\text{x}} \to 0} \dfrac{{{\text{2si}}{{\text{n}}^{\text{2}}}{\text{x}}}}{{{\text{1 - cosx}}}} $

$   = \;\mathop {\lim }\limits_{{\text{x}} \to 0} \dfrac{{{\text{2(1 - cosx)(1 + cosx)}}}}{{{\text{1 - cosx}}}} $

$   = \;\mathop {\lim }\limits_{{\text{x}} \to 0} {\text{2(1 + cosx)}} $

$  = \;2(1 + \cos 0) $

$   = \;2 \times 2\; = \;4 $

18. \[\mathop {\lim }\limits_{{\text{x}} \to 0} \dfrac{{{\text{ax + xcosx}}}}{{{\text{bsinx}}}}\]

उत्तर: \[\mathop {\lim }\limits_{{\text{x}} \to 0} \dfrac{{{\text{x(a + cosx)}}}}{{{\text{bsinx}}}}\]

\[\mathop {\lim }\limits_{{\text{x}} \to 0} (\dfrac{{\text{x}}}{{{\text{sinx}}}})\dfrac{{{\text{a + cosx}}}}{{\text{b}}}\]

\[\mathop {\lim }\limits_{{\text{x}} \to 0} \dfrac{{{\text{a + cosx}}}}{{\text{b}}}\]    ( \[\because \;\mathop {\lim }\limits_{{\text{x}} \to 0} \dfrac{{\text{x}}}{{{\text{sinx}}}}\; = \;1\] )

\[{\text{ = }}\;\dfrac{{{\text{a + cos0}}}}{{\text{b}}}\;{\text{ = }}\;\dfrac{{{\text{a + 1}}}}{{\text{b}}}\]

19. \[\mathop {\lim }\limits_{{\text{x}} \to 0} {\text{xsecx}}\]

उत्तर: \[\mathop {\lim }\limits_{{\text{x}} \to 0} (\dfrac{{\text{x}}}{{{\text{cosx}}}}{\text{)}}\]

\[ = \;\dfrac{0}{{{\text{cos0 }}}}\; = \;\dfrac{0}{1}\; = \;0\]

20. \[\mathop {\lim }\limits_{{\text{x}} \to 0} \dfrac{{{\text{sinax + bx}}}}{{{\text{ax + sinbx}}}},\;{\text{a,b,c}}\; \ne \;0\]

उत्तर: जब अंश और हर को \[{\text{x}}\] भाग कर के 

$   = \;\mathop {\lim }\limits_{{\text{x}} \to 0} \left( {\dfrac{{\dfrac{{\sin ax}}{x} + b}}{{a + \dfrac{{\sin bx}}{x}}}} \right) $

$  = \;\mathop {\lim }\limits_{x \to 0} \dfrac{{\left( {\dfrac{{{\text{sinax}}}}{{{\text{ax}}}}} \right){\text{ + a + b}}}}{{{\text{a + }}\left( {\dfrac{{{\text{sinbx}}}}{{{\text{bx}}}}} \right){\text{b}}}} $

\[ = \;\dfrac{{{\text{1a + b}}}}{{{\text{a + 1b}}}}\; = \;\dfrac{{{\text{a + b}}}}{{{\text{a + b}}}}\]

\[ = \;1,\;{\text{a + b}} \ne 0\]

21. \[\mathop {\lim }\limits_{{\text{x}} \to 0} {\text{(cosecx - cotx)}}\]

उत्तर: \[\mathop {\lim }\limits_{{\text{x}} \to 0} (\dfrac{{\text{1}}}{{{\text{sinx}}}}{\text{ - }}\dfrac{{{\text{cosx}}}}{{{\text{sinx}}}}{\text{)}}\]

$   = \;\mathop {\lim }\limits_{{\text{x}} \to 0} \dfrac{{{\text{(1 - cosx)} \times \text{sinx}}}}{{{\text{(sinx)(sinx)}}}} $

$   = \;\mathop {\lim }\limits_{{\text{x}} \to 0} \dfrac{{{\text{(1 - cosx)} \times \text{sinx}}}}{{{\text{(1 - cosx)(1 + cosx)}}}} $

$   = \;\mathop {\lim }\limits_{{\text{x}} \to 0} \left( {\dfrac{{{\text{sinx}}}}{{{\text{1 + cosx}}}}} \right) $

$   = \;\dfrac{0}{2}\; = \;0 $

22. \[\mathop {\lim }\limits_{{\text{x}} \to \dfrac{\pi }{2}} (\dfrac{{{\text{tan2x}}}}{{{\text{x - }}\dfrac{\pi }{2}}})\]

उत्तर: \[\mathop {\lim }\limits_{{\text{x}} \to \dfrac{\pi }{2}} (\dfrac{{{\text{tan2x}}}}{{{\text{x - }}\dfrac{\pi }{2}}})\] मे \[{\text{x}}\;{\text{ = }}\;\dfrac{{{\pi }}}{{\text{2}}}{\text{ + h}}\] रेखने पर 

 $   = \;\mathop {\lim }\limits_{{\text{h}} \to 0} \left( {\dfrac{{{\text{tan2}}\left( {\dfrac{{{\pi }}}{{\text{2}}}{\text{ + h}}} \right)}}{{\text{h}}}} \right) $

$   = \;\mathop {\lim }\limits_{{\text{h}} \to 0} \left( {\dfrac{{{{tan(\pi  + 2h)}}}}{{\text{h}}}} \right) $

$   = \;\mathop {\lim }\limits_{{\text{h}} \to 0} \left( {\dfrac{{{\text{tan(2h)}}}}{{\text{h}}}} \right) $

$   = \;\mathop {\lim }\limits_{{\text{h}} \to 0} \left( {\dfrac{{{\text{sin(2h)}}}}{{{\text{cos(2h)} \times \text{h}}}}} \right) $

$   = \;\mathop {\lim }\limits_{{\text{h}} \to 0} \left( {\dfrac{{{\text{sin(2h)} \times \text{ 2}}}}{{{\text{cos(2h)} \times \text{2h}}}}} \right) $

$  \left\{ {\because \;\mathop {\lim }\limits_{{\text{h}} \to 0} \left( {\dfrac{{{\text{sin(2h)}}}}{{{\text{2h}}}} = 1} \right)} \right\} $

$   = \;\mathop {\lim }\limits_{{\text{h}} \to 0} \left( {\dfrac{{\text{2}}}{{{\text{cos(2h)}}}}} \right) $

$   = \;2/\cos 0\; = \;2/1\; $

$   = \;2 $

23. \[\mathop {\lim }\limits_{{\text{x}} \to 0} {\text{f(x),}} और \;\mathop {\lim }\limits_{{\text{x}} \to 1} {\text{f(x)}}\] ज्ञात कीजिए जहा \[{\text{f(x)}} = \left\{ {\begin{array}{*{20}{r}} {{\text{2x + 3}},}&{{\text{x}} \leqslant 0} \\ {{\text{3(x + 1)}},}&{{\text{x}} \geqslant 0} \end{array}} \right.\]

उत्तर: 

(i) \[\mathop {\lim }\limits_{{\text{x}} \to 0} {\text{f(x)}}\]  के लिए सारणी इस प्रकार है 

\[\mathop {\lim }\limits_{{\text{x}} \to 0} {\text{f(x)}}\;{\text{ = }}\;{\text{3}}\]

\[{\text{f(x)}}\;{\text{ = }}\;{\text{3(x + 1)}}\]

\[\mathop {\lim }\limits_{{\text{x}} \to 0} {\text{f(x)}}\;{\text{ = }}\;{\text{3}}\]

\[{\text{f(x)}}\;{\text{ = }}\;{\text{3(x + 1)}}\]

\[{\text{x < 0,}}\;{\text{f(x)}}\;{\text{ = }}\;{\text{2x + 3}}\]

\[\mathop {\lim }\limits_{{\text{x}} \to {{\text{a}}^ - }} {\text{(2x + 3)}} = \mathop {\lim }\limits_{{\text{h}} \to 0} {\text{(2(0 - h) + 3) = 3}}\]

\[{\text{x > 0,}}\;{\text{f(x)}}\;{\text{ = }}\;{\text{3(1 + x)}}\]

\[\mathop {\lim }\limits_{{\text{x}} \to {{\text{a}}^ - }} {\text{(3x + 3)}} = \mathop {\lim }\limits_{{\text{h}} \to 0} {\text{(3(0 + h) + 3) = 3}}\]

(ii) \[\mathop {\lim }\limits_{{\text{x}} \to {1^ - }} {\text{f(x)}}\] के लिए सारणी इस प्रकार है 

\[{\text{x}}\;{\text{ = }}\;{\text{0}}{\text{.9999,}}\;{\text{f(x)}}\;{\text{ = }}\;{\text{5}}{\text{.9997}}\]

\[\mathop {\lim }\limits_{{\text{x}} \to {1^ - }} 3{\text{(x + 1)}}\;{\text{ = }}\;{\text{6}}\]

\[{\text{x}}\;{\text{ = }}\;1.0001{\text{,}}\;{\text{f(x)}}\;{\text{ = }}\;6.0003\]

\[\mathop {\lim }\limits_{{\text{x}} \to {1^ + }} 3{\text{(x + 1)}}\;{\text{ = }}\;{\text{6}}\]

\[\mathop {\lim }\limits_{{\text{x}} \to 1} {\text{f(x)}}\;{\text{ = }}\;{\text{6}}\]

24. \[\mathop {\lim }\limits_{{\text{x}} \to 1} {\text{f(x)}}\] जहा  \[{\text{f(x)}} = \left\{ {\begin{array}{*{20}{r}} {{{\text{x}}^{\text{2}}}{\text{ - 1}},}&{{\text{x}} \leqslant 1} \\ {{\text{ - }}{{\text{x}}^{\text{2}}}{\text{ - 1}},}&{{\text{x > 1}}} \end{array}} \right.\]

उत्तर: \[{\text{x < 1,}}\;{\text{f(x)}}\;{\text{ = }}\;{{\text{x}}^{\text{2}}}{\text{ - 1}}\]

 \[{\text{x}}\;{\text{ = }}\;0.999{\text{,}}\;{\text{f(x)}}\;{\text{ = }}\; - 0.001999\]

\[\mathop {\lim }\limits_{{\text{x}} \to {1^ - }} {\text{f(x)}}\;{\text{ = }}\;0\] ………………………..(i)

\[{\text{x > 1,}}\;{\text{f(x)}}\;{\text{ = }}\;{\text{ - }}{{\text{x}}^{\text{2}}}{\text{ - 1}}\]

\[{\text{x}}\;{\text{ = }}\;1.0001{\text{,}}\;{\text{f(x)}}\;{\text{ = }}\; - 2.002001\]

\[\mathop {\lim }\limits_{{\text{x}} \to  + 1} {\text{f(x)}}\;{\text{ = }}\; - 2\] ………………………(ii)

(i) और (ii) से \[\mathop {\lim }\limits_{{\text{x}} \to {1^ - }} {\text{f(x}}) \ne \mathop {\lim }\limits_{{\text{x}} \to  + 1} {\text{f(x)}}\]

इस प्रकार \[{\text{x}}\;{\text{ = }}\;{\text{1}}\] सीमा मोजूद नहीं है

25. \[\mathop {\lim }\limits_{{\text{x}} \to 0} {\text{f(x)}}\] जहा \[{\text{f(x)}} = \left\{ {\begin{array}{*{20}{r}} {|{\text{x}}|/{\text{x}},}&{{\text{x}} \ne 0} \\ {0,}&{{\text{x}} = 0} \end{array}} \right.\]

उत्तर: \[{\text{x}} < 0,\;\;{\text{|x|}}\;{\text{ = }}\;{\text{ - x}}\]

$  \mathop {\lim }\limits_{{\text{x}} \to {0^ - }} {\text{f(x}})\; = \;\mathop {\lim }\limits_{{\text{x}} \to {0^ - }} {\text{(|x|/x)}}\; = \;\mathop {\lim }\limits_{{\text{x}} \to {0^ - }} {\text{( - x/x)}}\; = \; - 1 $

$  {\text{x > 0,}}\;\;{\text{|x|}}\;{\text{ = }}\;{\text{x}} $

$  \mathop {\lim }\limits_{{\text{x}} \to  + 0} {\text{f(x}})\; = \;\mathop {\lim }\limits_{{\text{x}} \to  + 0} ({\text{|x|/x}})\; = \;\mathop {\lim }\limits_{{\text{x}} \to  + 0} ({\text{x/x)}}\; = \;1 $

इस प्रकार \[{\text{x}}\;{\text{ = }}\;0\] सीमा मोजूद नहीं है

26. \[\mathop {\lim }\limits_{{\text{x}} \to 0} {\text{f(x)}}\] जहा \[{\text{f(x)}} = \left\{ {\begin{array}{*{20}{r}} {|{\text{x}}|/{\text{x}},}&{{\text{x}} \ne 0} \\ {0,}&{{\text{x}} = 0} \end{array}} \right.\]

उत्तर: \[{\text{x}} < 0,\;\;{\text{|x|}}\;{\text{ = }}\;{\text{ - x}}\]

$  \mathop {\lim }\limits_{{\text{x}} \to {0^ - }} {\text{f(x}})\; = \;\mathop {\lim }\limits_{{\text{x}} \to {0^ - }} {\text{(|x|/x)}}\; = \;\mathop {\lim }\limits_{{\text{x}} \to {0^ - }} {\text{( - x/x)}}\; = \; - 1 $

$  {\text{x > 0,}}\;\;{\text{|x|}}\;{\text{ = }}\;{\text{x}} $

$  \mathop {\lim }\limits_{{\text{x}} \to  + 0} {\text{f(x}})\; = \;\mathop {\lim }\limits_{{\text{x}} \to  + 0} ({\text{|x|/x}})\; = \;\mathop {\lim }\limits_{{\text{x}} \to  + 0} ({\text{x/x)}}\; = \;1 $

इस प्रकार \[{\text{x}}\;{\text{ = }}\;0\] सीमा मोजूद नहीं है

27. \[\mathop {\lim }\limits_{{\text{x}} \to 5} {\text{f(x)}}\] जहा \[{\text{f(x)}}\;{\text{ = }}\;\left| {\text{x}} \right|{\text{ - 5}}\]

उत्तर: \[\mathop {\lim }\limits_{{\text{x}} \to {5^ - }} {\text{f(x)}}\;{\text{ = }}\;\mathop {\lim }\limits_{{\text{x}} \to {5^ - }} {\text{(}}\left| {\text{x}} \right|{\text{ - 5 )}}\]

$   = \;\mathop {\lim }\limits_{{\text{h}} \to 0} (|5 - {\text{h}}| - 5) $

$   = \;\mathop {\lim }\limits_{{\text{h}} \to 0} (5 - {\text{h}} - 5)\; = \;\mathop {\lim }\limits_{{\text{h}} \to 0} ( - {\text{h}})\; = \;0 $

$  \mathop {\lim }\limits_{{\text{x}} \to 5 + } {\text{f(x)}}\; = \;\mathop {\lim }\limits_{{\text{x}} \to 5 + } {\text{(|x| - 5)}}\; $

$   = \;\mathop {\lim }\limits_{{\text{h}} \to 0} (|5 + {\text{h}}| - 5) $

$   = \;\mathop {\lim }\limits_{{\text{h}} \to 0} (5 + {\text{h}} - 5)\; = \;\mathop {\lim }\limits_{{\text{h}} \to 0} (\;{\text{h}})\; = \;0 $

$  \mathop {\lim }\limits_{{\text{x}} \to {5^ - }} {\text{f(x)}}\; = \;\mathop {\lim }\limits_{{\text{x}} \to 5 + } {\text{f(x)}} $

28. मान लीजिए \[{\text{f(x)}} = \left\{ \begin{gathered} \begin{array}{*{20}{r}} {{\text{a + bx}},}&{{\text{x < }}1} \\ {4,}&{{\text{x = 1}}} \end{array} \hfill \\ {\text{b - ax,     x > 1}} \hfill \\ \end{gathered}  \right.\]

उत्तर: \[{\text{x < 1,}}\;{\text{f(x)}}\;{\text{ = }}\;{\text{a + bx}}\]

$  {\text{x}}\;{\text{ = }}\;{\text{0}}{\text{.999,}}\;{\text{f(x)}}\;{\text{ = }}\;{\text{a + 0}}{\text{.999b}} $

$  \mathop {\lim }\limits_{{\text{x}} \to {1^ - }} {\text{f(x)}}\;{\text{ = }}\;{\text{a + b}} $

\[{\text{x > 1,}}\;{\text{f(x)}}\;{\text{ = }}\;{\text{b - ax}}\]

$  {\text{x}}\;{\text{ = }}\;1.0001{\text{,}}\;{\text{f(x)}}\;{\text{ = }}\;{\text{b - 1}}{\text{.0001a}} $

$  \mathop {\lim }\limits_{{\text{x}} \to {1^ - }} {\text{f(x)}}\;{\text{ = }}\;{\text{a + b}} $

$  \mathop {\lim }\limits_{{\text{x}} \to 1 + } {\text{f(x)}}\;{\text{ = }}\;{\text{a + b}}\; = \;{\text{f(1)}}\;{\text{ = }}\;{\text{4}} $

\[{\text{a + b}}\;{\text{ = }}\;{\text{4}}\] …………….(i)

\[\mathop {\lim }\limits_{{\text{x}} \to {1^ - }} {\text{f(x)}}\;{\text{ = }}\;{\text{a + b}}\; = \;{\text{f(1)}}\;{\text{ = }}\;{\text{4}}\]

\[{\text{b - a}}\;{\text{ = }}\;{\text{4}}\] ………….(ii)

$  \therefore \;{\text{a + b + b - a}}\; = \;8 $

$  {\text{2b}}\;{\text{ = }}\;{\text{8}} $

$  {\text{b}}\;{\text{ = }}\;{\text{4,}}\;{\text{a}}\;{\text{ = }}\;{\text{0}} $

29. मान लीजिए \[{{\text{a}}_{\text{1}}}{\text{,}}{{\text{a}}_{\text{2}}}{\text{,}}......{\text{,}}{{\text{a}}_{\text{n}}}\] वास्तविक संख्या है और \[{\text{f(x)}}\;{\text{ = }}\;{\text{(x - }}{{\text{a}}_{\text{1}}}{\text{)(x - }}{{\text{a}}_{\text{2}}}{\text{)}}......{\text{(x - }}{{\text{a}}_{\text{n}}})\] से परिभाषा है। \[\mathop {\lim }\limits_{{\text{x}} \to {\text{a}}} {\text{f(x)}}\] क्या है। 

उत्तर: फैक्टर \[{\text{(x - }}{{\text{a}}_{\text{1}}}{\text{)}}\] के लिए 

यदि \[{\text{x > }}{{\text{a}}_{\text{1}}}{\text{,}}\;{\text{x - }}{{\text{a}}_{\text{1}}} > 0\]

$  \mathop {\lim }\limits_{{\text{x}} \to {a_1}} {\text{f}}\left( {{\text{x - }}{{\text{a}}_{\text{1}}}} \right)\;{\text{ = }}\;\left( {{{\text{a}}_{\text{1}}}{\text{ - }}{{\text{a}}_{\text{2}}}} \right) $

$  \mathop {\lim }\limits_{{\text{x}} \to {a_1}} {\text{f(x)}}\; = \;\left( {{\text{x - }}{{\text{a}}_{\text{1}}}} \right)\left( {{\text{x - }}{{\text{a}}_{\text{2}}}} \right){{ \dots }}\left( {{\text{x - }}{{\text{a}}_{\text{n}}}} \right) $

$   = \;\mathop {\lim }\limits_{x \to {a_1}} \left( {{\text{x - }}{{\text{a}}_{\text{1}}}} \right)\mathop {\lim }\limits_{{\text{x}} \to {a_1}} \left( {{\text{x - }}{{\text{a}}_{\text{2}}}} \right){{ \dots }}\left( {{\text{x - }}{{\text{a}}_{\text{n}}}} \right) $

$   = \;0 \times \mathop {\lim }\limits_{{\text{x}} \to {a_1}} \left( {{\text{x - }}{{\text{a}}_{\text{2}}}} \right){{ \dots }}\left( {{\text{x - }}{{\text{a}}_{\text{n}}}} \right)\; = \;0 $

जब \[{\text{a}}\;{\text{ = }}\;{{\text{a}}_{\text{1}}}{\text{,}}{{\text{a}}_{\text{2}}}{\text{,}}........{\text{,}}{{\text{a}}_{\text{n}}}\]

जैसे ही \[{\text{x > a,}}\;{\text{x - }}{{\text{a}}_{\text{1}}}{\text{ > a - }}{{\text{a}}_{\text{1}}}\]

\[{\text{a - }}{{\text{a}}_1}\] शून्य नहीं है 

इस प्रकार दूसरे कारक के मूल्य \[{\text{a - }}{{\text{a}}_{\text{1}}}{\text{,a - }}{{\text{a}}_{\text{2}}}{\text{,}}........{\text{,a - }}{{\text{a}}_{\text{n}}}\]

अब \[\mathop {\lim }\limits_{{\text{x}} \to {a_1}} {\text{f(x)}}\; = \;\left( {{\text{x - }}{{\text{a}}_{\text{1}}}} \right)\left( {{\text{x - }}{{\text{a}}_{\text{2}}}} \right){{ \dots }}\left( {{\text{x - }}{{\text{a}}_{\text{n}}}} \right)\]

\[ = \;\left( {{\text{a - }}{{\text{a}}_{\text{1}}}} \right)\left( {{\text{a - }}{{\text{a}}_{\text{2}}}} \right){{ \dots }}\left( {{\text{a - }}{{\text{a}}_{\text{n}}}} \right)\]

30. यदि \[{\text{f(x)}} = \left\{ \begin{gathered} \begin{array}{*{20}{r}} {\left| {\text{x}} \right| + 1,}&{{\text{x < }}1} \\ {0,}&{{\text{x = 1}}} \end{array} \hfill \\ \left| {\text{x}} \right|{\text{ - 1  ,     x > 1}} \hfill \\ \end{gathered}  \right.\]

उत्तर: (i) \[{\text{x}}\;{\text{ = }}\;{\text{0}}\] ,

$  \mathop {\lim }\limits_{{\text{x}} \to {0^ - }} {\text{f(x)}}\; = \;\mathop {\lim }\limits_{{\text{x}} \to {0^ - }} {\text{(1 - x)}}\; = \;1 $

$  \mathop {\lim }\limits_{{\text{x}} \to 0 + } {\text{f(x)}}\; = \;\mathop {\lim }\limits_{{\text{x}} \to 0 + } {\text{(x - 1)}}\; = \; - 1 $

$  \mathop {\lim }\limits_{{\text{x}} \to {0^ - }} {\text{f(x)}}\; \ne \;\mathop {\lim }\limits_{{\text{x}} \to 0 + } {\text{f(x)}} $

 \[{\text{x}}\;{\text{ = }}\;{\text{0}}\] पर \[\mathop {\lim }\limits_{{\text{x}} \to 0} {\text{f(x)}}\] मोजूद नहीं है

(ii) \[{\text{a < 0}}\] ,

$  \mathop {\lim }\limits_{{\text{x}} \to {{\text{a}}^{\text{ - }}}} {\text{f(x)}}\; = \;\mathop {\lim }\limits_{{\text{x}} \to {{\text{a}}^{\text{ - }}}} {\text{(1 - x)}}\;{\text{ = }}\;{\text{1 - a}} $

$  \mathop {\lim }\limits_{{\text{x}} \to {\text{a}} + } {\text{f(x)}}\; = \;\mathop {\lim }\limits_{{\text{x}} \to {\text{a + }}} {\text{(1 - x)}}\;{\text{ = }}\;{\text{1 - a}} $

$  \mathop {\lim }\limits_{{\text{x}} \to {{\text{a}}^{\text{ - }}}} {\text{f(x)}}\; = \;\mathop {\lim }\limits_{{\text{x}} \to {\text{a + }}} {\text{f(x)}} $

$  \mathop {\lim }\limits_{{\text{x}} \to {\text{a}}} {\text{f(x)}}\; = \;{\text{1 - a}} $

(iii) \[{\text{a > 0}}\] ,

$  \mathop {\lim }\limits_{{\text{x}} \to {{\text{a}}^{\text{ - }}}} {\text{f(x)}}\; = \;\mathop {\lim }\limits_{{\text{x}} \to {{\text{a}}^{\text{ - }}}} {\text{(x - 1)}}\;{\text{ = }}\;{\text{a - 1}} $

$ \mathop {\lim }\limits_{{\text{x}} \to {\text{a}} + } {\text{f(x)}}\; = \;\mathop {\lim }\limits_{{\text{x}} \to {\text{a + }}} {\text{(x - 1)}}\;{\text{ = }}\;{\text{a - 1}} $

$\mathop {\lim }\limits_{{\text{x}} \to {{\text{a}}^{\text{ - }}}} {\text{f(x)}}\; = \;\mathop {\lim }\limits_{{\text{x}} \to {\text{a + }}} {\text{f(x)}} $

$  \mathop {\lim }\limits_{{\text{x}} \to {\text{a}}} {\text{f(x)}}\; = \;{\text{a - 1}} $

इस तरह 

जब \[{\text{a < 0}}\] , \[\mathop {\lim }\limits_{{\text{x}} \to {\text{a}}} {\text{f(x)}}\;{\text{ = }}\;1 - {\text{a }}\]

जब \[{\text{a > 0}}\] , \[\mathop {\lim }\limits_{{\text{x}} \to {\text{a}}} {\text{f(x)}}\;{\text{ = }}\;{\text{a - 1 }}\]

जब \[{\text{a}} \ne 0\] के लिए \[\mathop {\lim }\limits_{{\text{x}} \to {\text{a}}} {\text{f(x)}}\] का अस्तित्व है।

31. यदि फलन \[{\text{f(x)}}\] , \[\mathop {\lim }\limits_{{\text{x}} \to 1} {\text{(f(x) - 2)/}}{{\text{x}}^2} - 1\; = \;\pi \] को संतुष्ट करता है तो \[\mathop {\lim }\limits_{{\text{x}} \to 1} {\text{ f(x)}}\] का मान प्राप्त कीजिए। 

उत्तर: जैसे ही \[{\text{x}}\;{\text{ = }}\;{\text{1}}\] , \[\mathop {\lim }\limits_{{\text{x}} \to 1} {\text{(f(x) - 2)/}}{{\text{x}}^2} - 1\; = \;\pi \] दिया हुआ है 

जैसे ही \[{\text{x}}\;{\text{ = }}\;{\text{1}}\] , \[{{\text{x}}^2}{\text{ - 1}}\;{\text{ = }}\;{\text{0 }}\]

जिससे, जैसे ही \[{\text{x}}\;{\text{ = }}\;{\text{1}}\] , \[\dfrac{{{\text{f(x) - 2 }}}}{{{{\text{x}}^2}{\text{ - 1}}}}\; = \;\dfrac{0}{0}\] रूप मे होगा 

जैसे ही \[{\text{x}}\;{\text{ = }}\;{\text{1}}\] , \[{\text{f(x) - 2}}\; = \;0\]

\[\mathop {\lim }\limits_{{\text{x}} \to 1} {\text{f(x)}}\;{\text{ = }}\;{\text{2 }}\]

32. यदि सीमा \[\mathop {\lim }\limits_{{\text{x}} \to 0} {\text{ f(x)}}\] और \[\mathop {\lim }\limits_{{\text{x}} \to 1} {\text{ f(x)}}\] मोजूद है तो \[{\text{m}}\] और \[{\text{n}}\] के मान ज्ञात करे 

\[{\text{f(x)}} = \left\{ {\begin{array}{*{20}{c}} {{\text{m}}{{\text{x}}^{\text{2}}}{\text{ + n}},}&{{\text{x}} < 0} \\ {{\text{nx + m}},}&{0 \leqslant {\text{x}} \leqslant 1} \\ {{\text{n}}{{\text{x}}^{\text{3}}}{\text{ + m}},}&{{\text{x}} > 1} \end{array}} \right.\]

उत्तर: जब \[{\text{x}}\;{\text{ = }}\;{\text{0}}\] , 

$  \mathop {\lim }\limits_{{\text{x}} \to {0^ - }} {\text{f(x)}}\; = \;\mathop {\lim }\limits_{{\text{x}} \to {0^ - }} {\text{f}}\left( {{\text{m}}{{\text{x}}^{\text{2}}}{\text{ + n}}} \right)\;{\text{ = }}\;{\text{n}} $

$  \mathop {\lim }\limits_{{\text{x}} \to 0 + } {\text{f(x)}}\; = \;\mathop {\lim }\limits_{{\text{x}} \to 0 + } {\text{f(nx + m)}}\;{\text{ = }}\;{\text{m}} $

जब \[{\text{x}}\;{\text{ = }}\;1\] ,

$  \mathop {\lim }\limits_{{\text{x}} \to {1^ - }} {\text{f(x)}}\; = \;\mathop {\lim }\limits_{{\text{x}} \to {1^ - }} {\text{f}}\left( {{\text{nx + m}}} \right)\;{\text{ = }}\;{\text{m + n}} $

$  \mathop {\lim }\limits_{{\text{x}} \to 1 + } {\text{f(x)}}\; = \;\mathop {\lim }\limits_{{\text{x}} \to 1 + } {\text{f(nx + m)}}\;{\text{ = }}\;{\text{m + n }} $

\[\mathop {\lim }\limits_{{\text{x}} \to 0} {\text{ f(x)}}\] के अस्तित्व के लिए \[{\text{m}}\] और \[{\text{n}}\] समान होना चाहिए और \[\mathop {\lim }\limits_{{\text{x}} \to 1} {\text{ f(x)}}\] के अस्तित्व के लिए \[{\text{m}}\] और \[{\text{n}}\] कोई भी पूर्णक मूल्य होना चाहिए।

प्रश्नावली 13.2

1. \[{\text{x}}\;{\text{ = }}\;{\text{10}}\] पर \[{{\text{x}}^2} - 2\] का अवकलज ज्ञात कीजिए। 

उत्तर: \[{\text{x}}\;{\text{ = }}\;{\text{a}}\] पर \[{\text{f(x)}}\] का अवकलज 

= \[\mathop {\lim }\limits_{{\text{h}} \to 0} \dfrac{{{\text{f(a + h) - f(a)}}}}{{\text{h}}}\]

\[{\text{x}}\;{\text{ = }}\;{\text{10}}\] पर \[{{\text{x}}^2} - 2\] का अवकलज,

$\text{f}^\prime(a)$=$ \;\mathop {\lim }\limits_{{\text{h}} \to 0} \dfrac{{\left[ {{{{\text{(10 + h)}}}^{\text{2}}}{\text{ - 2}}} \right]{\text{ - }}\left( {{\text{1}}{{\text{0}}^{\text{2}}}{\text{ - 2}}} \right)}}{{\text{h}}}$

$   = \;\mathop {\lim }\limits_{{\text{h}} \to 0} \dfrac{{{\text{100 + 20h + }}{{\text{h}}^{\text{2}}}{\text{ - 2 - 100 + 2}}}}{{\text{h}}} $

$   = \;\mathop {\lim }\limits_{{\text{h}} \to 0} \dfrac{{{\text{20h + }}{{\text{h}}^{\text{2}}}}}{{\text{h}}} $

$   = \;\mathop {\lim }\limits_{{\text{h}} \to 0} {\text{(20 + h)}} $

$   = \;20 $

2. \[{\text{x}}\;{\text{ = }}\;{\text{1}}\] पर \[{\text{x}}\] का अवकलज ज्ञात कीजिए। 

उत्तर: \[{\text{x}}\;{\text{ = }}\;{\text{a}}\] पर \[{\text{f(x)}}\] का अवकलज = \[\mathop {\lim }\limits_{{\text{h}} \to 0} \dfrac{{{\text{f(a + h) - f(a)}}}}{{\text{h}}}\]

\[{\text{x}}\;{\text{ = }}\;{\text{1}}\] पर \[{\text{x}}\] का अवकलज,

\[\text{f}^\prime(a)\; = \;\mathop {\lim }\limits_{{\text{h}} \to 0} \dfrac{{(1 + {\text{h) - 1}}}}{{\text{h}}} = \;1\]

3. \[{\text{x}}\;{\text{ = }}\;{\text{100 }}\] पर \[{\text{99x}}\] का अवकलज ज्ञात कीजिए। 

उत्तर: \[{\text{x}}\;{\text{ = }}\;{\text{a}}\] पर \[{\text{f(x)}}\] का अवकलज = \[\mathop {\lim }\limits_{{\text{h}} \to 0} \dfrac{{{\text{f(a + h) - f(a)}}}}{{\text{h}}}\]

\[{\text{x}}\;{\text{ = }}\;{\text{100 }}\] पर \[{\text{99x}}\] का अवकलज,

$  \text{f}^\prime(a)\; = \;\mathop {\lim }\limits_{{\text{h}} \to 0} \dfrac{{99(100 + {\text{h) - 99}} \times {\text{100}}}}{{\text{h}}} = \;\mathop {\lim }\limits_{{\text{h}} \to 0} \dfrac{{{{99 \times h}}}}{{\text{h}}} $

$   = \;99 $

4. प्रथम सिद्धांतसे निम्नलिखित फलनों के अवकलज ज्ञात कीजिए: 

(i) \[{{\text{x}}^3} - 27\]

उत्तर: \[{\text{f(x)}}\] = \[{{\text{x}}^3} - 27\]

\[{\text{f(x + h)}}\] = \[{{\text{(x + h)}}^3} - 27\; = \;{{\text{x}}^{\text{3}}}{\text{ + 3}}{{\text{x}}^{\text{2}}}{\text{h + 3x}}{{\text{h}}^{\text{2}}}{\text{ + }}{{\text{h}}^{\text{3}}}{\text{ - 27}}\]

\[{\text{f(x + h) - f(x)}}\] = \[({{\text{x}}^{\text{3}}}{\text{ + 3}}{{\text{x}}^{\text{2}}}{\text{h + 3x}}{{\text{h}}^{\text{2}}}{\text{ + }}{{\text{h}}^{\text{3}}}{\text{ - 27) - (}}{{\text{x}}^3} - 27)\]

= \[{\text{3}}{{\text{x}}^{\text{2}}}{\text{ + 3x}}{{\text{h}}^{\text{2}}}{\text{ + }}{{\text{h}}^{\text{3}}}\;{\text{ = }}\;{\text{h(3}}{{\text{x}}^{\text{2}}}{\text{ + 3xh + }}{{\text{h}}^{\text{2}}}{\text{)}}\]

\[{{\text{f}}^\prime}{\text{(x)}}\; = \;\mathop {\lim }\limits_{{\text{h}} \to 0} \dfrac{{{\text{f(x + h) - f(x)}}}}{{\text{h}}}\]

$ = \;\mathop {\lim }\limits_{{\text{h}} \to 1} \dfrac{{{\text{h}}\left( {{\text{3}}{{\text{x}}^{\text{2}}}{\text{h + 3}}{{\text{x}}^{\text{2}}}{\text{h + }}{{\text{h}}^{\text{3}}}} \right)}}{{\text{h}}}$

$   = \;{\text{3}}{{\text{x}}^{\text{2}}} $

(ii) \[{\text{(x - 1)(x - 2)}}\]

उत्तर: \[{\text{f(x)  =  (x - 1)(x - 2)  =  }}{{\text{x}}^{\text{2}}}{\text{ - 3x + 2}}\]

$ f(x + h) = {x + h}^2 - 3(x + h) + 2 $

$   = \;\left( {{{\text{x}}^{\text{2}}}{\text{ + 2xh + }}{{\text{h}}^{\text{2}}}} \right) - {\text{(3x + 3h)}} + 2 $

$  {\text{f(x + h) - f(x)}} = \;\left( {{{\text{x}}^{\text{2}}}{\text{ + 2xh + }}{{\text{h}}^{\text{2}}}} \right) - {\text{(3x + 3h)}} + 2 - \left( {{{\text{x}}^{\text{2}}}{\text{ - 3x + 2}}} \right) $

$   = \;{\text{2xh + }}{{\text{h}}^{\text{2}}}{\text{ - 3h}} $

$   = \;{\text{h(2x + h - 3)}} $

$ {{\text{f}}^\prime}{\text{(x)}} = \;\mathop {\lim }\limits_{{\text{h}} \to 0} \dfrac{{{\text{f(x + h) - f(x)}}}}{{\text{h}}} $

$   = \;\mathop {\lim }\limits_{{\text{h}} \to 0} \dfrac{{{\text{h(2x + h - 3)}}}}{{\text{h}}} $

$   = \;{\text{2x - 3}} $

(iii) \[\dfrac{{\text{1}}}{{{{\text{x}}^2}}}\]

उत्तर: \[{\text{f(x)}}\;{\text{ = }}\;\dfrac{{\text{1}}}{{{{\text{x}}^{\text{2}}}}}\]

\[{\text{f(x + h)}}\;{\text{ = }}\;\dfrac{{\text{1}}}{{{{{\text{(x + h)}}}^{\text{2}}}}}\]

\[{\text{f(x + h) - f(x)}}\;{\text{ = }}\;\dfrac{{\text{1}}}{{{{{\text{(x + h)}}}^{\text{2}}}}}{\text{ - }}\dfrac{{\text{1}}}{{{{\text{x}}^{\text{2}}}}}\]

= $  \dfrac{{{{\text{x}}^{\text{2}}}{\text{ - (x + h}}{{\text{)}}^{\text{2}}}}}{{{{\text{x}}^{\text{2}}}{{{\text{(x + h)}}}^{\text{2}}}}} $

$   = \;\dfrac{{{{\text{x}}^{\text{2}}}{\text{ - }}\left[ {{{\text{x}}^{\text{2}}}{\text{ + 2xh + }}{{\text{h}}^{\text{2}}}} \right]}}{{{{\text{x}}^{\text{2}}}{{{\text{(x + h)}}}^{\text{2}}}}} $

$   = \;\dfrac{{{\text{ - h(2x + h)}}}}{{{{\text{x}}^{\text{2}}}{{{\text{(x + h)}}}^{\text{2}}}}} $

 \[{{\text{f}}^\prime}{\text{(x)}}\; = \;\mathop {\lim }\limits_{{\text{h}} \to 0} \dfrac{{{\text{f(x + h) - f(x)}}}}{{\text{h}}}\]

= \[\mathop {\lim }\limits_{{\text{h}} \to 0} \dfrac{{{\text{ - h(2x + h)}}}}{{{{\text{x}}^{\text{2}}}{{{\text{(x + h)}}}^{\text{2}}}{\text{h}}}}\; = \;\dfrac{{{\text{ - 2x}}}}{{{{\text{x}}^{\text{2}}}{{\text{x}}^{\text{2}}}}}\; = \;\dfrac{{{\text{ - 2}}}}{{{{\text{x}}^{\text{3}}}}}\]

(iv) \[\dfrac{{{\text{x + 1}}}}{{{\text{x - 1}}}}\]

उत्तर: \[{\text{f(x)}}\;{\text{ = }}\;\dfrac{{{\text{x + 1}}}}{{{\text{x - 1}}}}\]

\[{\text{f(x + h)}}\;{\text{ = }}\;\dfrac{{{\text{x + h + 1}}}}{{{\text{x + h - 1}}}}\]

\[{\text{f(x + h) - f(x)}}\;{\text{ = }}\;\dfrac{{{\text{x + h + 1}}}}{{{\text{x + h - 1}}}}{\text{ - }}\dfrac{{{\text{x + 1}}}}{{{\text{x - 1}}}}\;{\text{ = }}\;\dfrac{{{\text{ - 2h}}}}{{{\text{(x - 1)(x - 1 + h)}}}}\]

\[{{\text{f}}^\prime}{\text{(x)}}\; = \;\mathop {\lim }\limits_{{\text{h}} \to 0} \dfrac{{{\text{f(x + h) - f(x)}}}}{{\text{h}}}\]

 = \[\mathop {\lim }\limits_{{\text{h}} \to 0} \dfrac{{{\text{ - 2h}}}}{{{\text{(x - 1)(x - 1 + h)h}}}}\; = \;\dfrac{{{\text{ - 2}}}}{{{{{\text{(x - 1)}}}^2}}}\]

5. फलन \[{\text{f(x)}}\;{\text{ = }}\;\dfrac{{{{\text{x}}^{{\text{100}}}}}}{{{\text{100}}}}{\text{ + }}\dfrac{{{{\text{x}}^{{\text{99}}}}}}{{{\text{99}}}}{{ +  \dots }}..{\text{ + }}\dfrac{{{{\text{x}}^{\text{2}}}}}{{\text{2}}}{\text{ + x + 1}}\] के लिए सिद्ध कीजिए की \[{{\text{f}}^\prime}{\text{(1)}}\;{\text{ = }}\;{\text{100}}{{\text{f}}^\prime}{\text{(0)}}\]

उत्तर: \[\dfrac{{\text{d}}}{{{\text{dx}}}}{{\text{x}}^{\text{n}}}\;{\text{ = }}\;{\text{n}}{{\text{x}}^{{\text{n - 1}}}}\]

\[\text{f(x)}\text{ = }\dfrac{x^{100}}{100}+\dfrac{x^{99}}{99} +  \dots..{\text{ + }}\dfrac{{{{\text{x}}^{\text{2}}}}}{{\text{2}}}{\text{ + x + 1}}\]

$  {{\text{f}}^\prime}{\text{(x)}}\;{\text{ = }}\;\dfrac{{{\text{100}}{{\text{x}}^{99}}}}{{{\text{100}}}}{\text{ + }}\dfrac{{{\text{99}}{{\text{x}}^{{\text{98}}}}}}{{{\text{99}}}}{{ +  \dots }}..{\text{ + }}\dfrac{{{\text{2x}}}}{{\text{2}}}{\text{ + 1}} $

$  {\text{ = }}\;{{\text{x}}^{{\text{99}}}}{\text{ + }}{{\text{x}}^{{\text{98}}}}{\text{ + }}....{\text{ + x + 1}} $

जब \[{\text{x}}\;{\text{ = }}\;{\text{1}}\] , \[{{\text{f}}^\prime}(1)\; = \;1 + 1 + .... + 1\]

जब \[{\text{x}}\;{\text{ = }}\;0\] , \[{{\text{f}}^\prime}(0)\; = \;1\]

\[{{\text{f}}^\prime}(1)\; = \;100\]

\[{\text{100}}{{\text{f}}^\prime}(0)\; = \;100 \times 1\; = \;100\] 

अतः बाया पक्ष = दाया पक्ष

6. किसी आचार वास्तविक संख्या \[{\text{a}}\] के लिए \[{{\text{x}}^{\text{n}}}{\text{ + a}}{{\text{x}}^{{\text{n - 1}}}}{\text{ + }}{{\text{a}}^{\text{2}}}{{\text{x}}^{{\text{n - 2}}}}{{ +  \dots }}..{\text{ + }}{{\text{a}}^{{\text{n - 1}}}}{\text{x + }}{{\text{a}}^{\text{n}}}\] का अवकलज ज्ञात कीजिए। 

उत्तर: \[\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{[f(x)]}}\; = \;{{\text{f}}^\prime}{\text{(x)}}\]

$  \dfrac{{\text{d}}}{{{\text{dx}}}}{{\text{x}}^{\text{n}}}{\text{ = n}}{{\text{x}}^{{\text{n - 1}}}} $

$  {\text{f(x)}}\; = \;{{\text{x}}^{\text{n}}}{\text{ + a}}{{\text{x}}^{{\text{n - 1}}}}{\text{ + }}{{\text{a}}^{\text{2}}}{{\text{x}}^{{\text{n - 2}}}}{\text{ + }}......{\text{ + }}{{\text{a}}^{{\text{n - 1}}}}{\text{x + }}{{\text{a}}^{\text{n}}} $

$  {{\text{f}}^\prime }({\text{x}})\; = \;{\text{n}}{{\text{x}}^{{\text{n - 1}}}}{\text{ + (n - 1)a}}{{\text{x}}^{{\text{n - 2}}}}{\text{ + (n - 2)}}{{\text{a}}^{\text{2}}}{{\text{x}}^{{\text{n - 3}}}}{\text{ + }}......{\text{ + }}{{\text{a}}^{{\text{n - 1}}}} $

7. किन्ही अचरों \[{\text{a,}}\;{\text{b }}\] के लिए, अवकलज ज्ञात कीजिए:

(i) \[{\text{(x - a)(x - b)}}\]

उत्तर: \[{\text{f(x)}}\;{\text{ = }}\;{\text{(x - a)(x - b)}}\;{\text{ = }}\;{{\text{x}}^{\text{2}}}{\text{ - (a + b)x + ab}}\]

$  {{\text{f}}^\prime}{\text{(x)}}\;{\text{ = }}\;2{{\text{x}}^{{\text{2 - 1}}}}{\text{ - (a + b) + ab}} $

$  {\text{ = }}\;{\text{2x - (a + b)}} $

(ii) \[{{\text{(a}}{{\text{x}}^{\text{2}}}{\text{ + b)}}^{\text{2}}}\]

उत्तर: \[{\text{f(x)}}\;{\text{ = }}\;{{\text{(a}}{{\text{x}}^{\text{2}}}{\text{ + b)}}^{\text{2}}}\;{\text{ = }}\;{{\text{a}}^{\text{2}}}{{\text{x}}^{\text{4}}}{\text{ + 2ab}}{{\text{x}}^{\text{2}}}{\text{ + }}{{\text{b}}^{\text{2}}}\]

$  {{\text{f}}^\prime }{\text{(x)}}\; = \;{{\text{a}}^{\text{2}}}\left( {{\text{4}}{{\text{x}}^{\text{3}}}} \right){\text{ + 2ab(2x) + 0}} $

$   = \;{\text{4ax}}\left( {{\text{a}}{{\text{x}}^{\text{2}}}{\text{ + b}}} \right) $

(iii) \[\dfrac{{{\text{x - a}}}}{{{\text{x - b }}}}\]

उत्तर: \[{\text{f(x)}}\;{\text{ = }}\;\dfrac{{{\text{x - a}}}}{{{\text{x - b}}}}\;{\text{ = }}\;\dfrac{{\text{u}}}{{\text{v}}}\]

$  {\left( {\dfrac{{\text{u}}}{{\text{v}}}} \right)^\prime }\;{\text{ = }}\;\dfrac{{{{\text{u}}^\prime}{\text{v - u}}{{\text{v}}^\prime}}}{{{{\text{v}}^{\text{2}}}}} $

$  {{\text{f}}^\prime }({\text{x)}}\; = \;\;\dfrac{{{\text{1(x - b) - (x - a)1}}}}{{{{{\text{(x - b)}}}^{\text{2}}}}} $

$  {\text{ = }}\;\dfrac{{{\text{a - b}}}}{{{{{\text{(x - b)}}}^{\text{2}}}}} $

8. किसी आचार \[{\text{a}}\] के लिए \[\dfrac{{{{\text{x}}^{\text{n}}}{\text{ - }}{{\text{a}}^{\text{n}}}}}{{{\text{x - a}}}}\] का अवकलज ज्ञात कीजिए। 

उत्तर:  \[{\text{f(x)}}\;{\text{ = }}\;\dfrac{{{{\text{x}}^{\text{n}}}{\text{ - }}{{\text{a}}^{\text{n}}}}}{{{\text{x - a}}}}\;{\text{ = }}\;\dfrac{{\text{u}}}{{\text{v}}}\]

$  {\left( {\dfrac{{\text{u}}}{{\text{v}}}} \right)^\prime }\; = \;\dfrac{{{{\text{u}}^\prime}{\text{v - u}}{{\text{v}}^\prime}}}{{{{\text{v}}^{\text{2}}}}}{{\text{f}}^\prime }({\text{x}})\; $

$   = \;\dfrac{{{\text{n}}{{\text{x}}^{{\text{n - 1}}}}{\text{(x - a) - }}\left( {{{\text{x}}^{\text{n}}}{\text{ - }}{{\text{a}}^{\text{n}}}} \right){{ \times 1}}}}{{{{{\text{(x - a)}}}^{\text{2}}}}} $

$  {\text{ = }}\;\dfrac{{{\text{n}}{{\text{x}}^{\text{n}}}{\text{ - na}}{{\text{x}}^{{\text{n - 1}}}}{\text{ - }}{{\text{x}}^{\text{n}}}{\text{ + }}{{\text{a}}^{\text{n}}}}}{{{{{\text{(x - a)}}}^{\text{2}}}}} $

9. निम्नलिखित के अवकलज ज्ञात कीजिए: 

(i) \[{\text{2x - }}\dfrac{3}{4}{\text{ }}\]

उत्तर: \[{\text{f(x)}}\;{\text{ = }}\;{\text{2x - }}\dfrac{{\text{3}}}{{\text{4}}}\]

\[{{\text{f}}^\prime}{\text{(x)}}\;{\text{ = }}\;{\text{2(1) - 0}}\;{\text{ = }}\;{\text{2}}\]

(ii) \[{\text{(5}}{{\text{x}}^{\text{3}}}{\text{ + 3x + 1)(x - 1)}}\]

उत्तर: \[{\text{f(x)}}\] = \[{\text{(5}}{{\text{x}}^{\text{3}}}{\text{ + 3x + 1)(x - 1)}}\]

$  \dfrac{{\text{d}}}{{{\text{dx}}}}{\text{(uv}})\; = \;{{\text{u}}^\prime }{\text{v}} + {\text{u}}{{\text{v}}^\prime } $

$  {{\text{f}}^\prime }({\text{x}})\; = \;\left( {{\text{15}}{{\text{x}}^{\text{2}}}{\text{ + 3}}} \right){\text{(x - 1) + }}\left( {{\text{5}}{{\text{x}}^{\text{3}}}{\text{ + 3x + 1}}} \right){\text{(1)}} $

$ = \;{\text{20}}{{\text{x}}^{\text{3}}}{\text{ - 15}}{{\text{x}}^{\text{2}}}{\text{ + 6x - 4}} $

(iii) \[{{\text{x}}^{ - 3}}{\text{(5 + 3x)}}\]

उत्तर: \[{\text{f(x)}}\] = \[{{\text{x}}^{ - 3}}{\text{(5 + 3x)}}\]

$  \dfrac{{\text{d}}}{{{\text{dx}}}}{\text{(uv}})\; = \;{{\text{u}}^\prime }{\text{v}} + {\text{u}}{{\text{v}}^\prime } $

$  {{\text{f}}^\prime }({\text{x}})\; = \;{\text{5( - 3)}}{{\text{x}}^{{\text{ - 3 - 1}}}}{\text{ + 3( - 2)}}{{\text{x}}^{{\text{ - 2 - 1}}}} $

$= \;{\text{ - 15}}{{\text{x}}^{{\text{ - 4}}}}{\text{ - 6}}{{\text{x}}^{{\text{ - 3}}}}$

$   = \;\dfrac{{{\text{ - 15}}}}{{{{\text{x}}^{\text{4}}}}}{\text{ - }}\dfrac{{\text{6}}}{{{{\text{x}}^{\text{3}}}}}{\text{ - }}\dfrac{{\text{3}}}{{{{\text{x}}^{\text{4}}}}}{\text{(5 + 2x)}} $

(iv) \[{{\text{x}}^5}{\text{(3 - 6}}{{\text{x}}^{ - 9}}{\text{)}}\]

उत्तर: \[{\text{f(x)}}\] = \[{{\text{x}}^5}{\text{(3 - 6}}{{\text{x}}^{ - 9}}{\text{)}}\] = \[{\text{3}}{{\text{x}}^{\text{5}}}{\text{ - 6}}{{\text{x}}^{{\text{ - 4}}}}\]

$  \dfrac{{\text{d}}}{{{\text{dx}}}}{\text{(uv)}}\; = \;{{\text{u}}^\prime }{\text{v}} + {\text{u}}{{\text{v}}^\prime } $

$  {{\text{f}}^\prime }({\text{x}})\; = \;1{\text{5}}{{\text{x}}^4}{\text{ + 24}}{{\text{x}}^{ - 5}} $

$   = \;{\text{15}}{{\text{x}}^{\text{4}}} + \dfrac{{24}}{{{{\text{x}}^5}}} $

(v) \[{{\text{x}}^{ - 4}}{\text{(3 - 4}}{{\text{x}}^{ - 5}}{\text{)}}\]

उत्तर: \[{\text{f(x)}}\] = \[{{\text{x}}^{ - 4}}{\text{(3 - 4}}{{\text{x}}^{ - 5}}{\text{)}}\] = \[{\text{3}}{{\text{x}}^{ - 4}}{\text{ - 4}}{{\text{x}}^{{\text{ - 9}}}}\]

$  \dfrac{{\text{d}}}{{{\text{dx}}}}{\text{(uv)}}\; = \;{{\text{u}}^\prime }{\text{v}} + {\text{u}}{{\text{v}}^\prime } $

$  {{\text{f}}^\prime }({\text{x}})\; = \;{\text{ - 12}}{{\text{x}}^{{\text{ - 5}}}}{\text{ + 36}}{{\text{x}}^{{\text{ - 10}}}} $

$   = \;{\text{ - }}\dfrac{{{\text{12}}}}{{{{\text{x}}^{\text{5}}}}}{\text{ + }}\dfrac{{{\text{36}}}}{{{{\text{x}}^{{\text{10}}}}}} $

(vi) \[\dfrac{{\text{2}}}{{{\text{x + 1}}}}{\text{ - }}\dfrac{{{{\text{x}}^{\text{2}}}}}{{{\text{3x - 1}}}}\]

उत्तर: \[{\text{f(x)}}\] = \[\dfrac{{\text{2}}}{{{\text{x + 1}}}}{\text{ - }}\dfrac{{{{\text{x}}^{\text{2}}}}}{{{\text{3x - 1}}}}\]

$  {\left( {\dfrac{{\text{u}}}{{\text{v}}}} \right)^\prime }\; = \;\dfrac{{{{\text{u}}^\prime}{\text{v - u}}{{\text{v}}^\prime}}}{{{{\text{v}}^{\text{2}}}}} $

$  {{\text{f}}^\prime }{\text{(x)}}\; = \;\dfrac{{{\text{(0 - 2)} \times 1}}}{{{{{\text{(x + 1)}}}^{\text{2}}}}}{\text{ - }}\dfrac{{{\text{2x(3x - 1) - }}{{\text{x}}^{\text{2}}}{{ \times\; 3}}}}{{{{{\text{(3x - 1)}}}^{\text{2}}}}} $

$  {\text{ = }}\;\dfrac{{{\text{ - 2}}}}{{{{{\text{(x + 1)}}}^{\text{2}}}}}{\text{ - }}\dfrac{{{\text{6}}{{\text{x}}^{\text{2}}}{\text{ - 2x - 3}}{{\text{x}}^{\text{2}}}}}{{{{{\text{(3x - 1)}}}^{\text{2}}}}} $

$  {\text{ = }}\;\dfrac{{{\text{ - 2}}}}{{{{{\text{(x + 1)}}}^{\text{2}}}}}{\text{ - }}\dfrac{{{\text{3}}{{\text{x}}^{\text{2}}}{\text{ - 2x}}}}{{{{{\text{(3x - 1)}}}^{\text{2}}}}} $

10. प्रथम सिद्धांत से \[{\text{cosx}}\] का अवकलज कीजिए। 

उत्तर: \[{\text{f(x)}}\] = \[{\text{cosx}}\]

\[{\text{f(x + h)}}\;{\text{ = }}\;{\text{cos(x + h)}}\]

$   = \;{\text{ - 2sin}}\dfrac{{{\text{x + h + x}}}}{{\text{2}}}{\text{sin}}\dfrac{{{\text{x + h - x}}}}{{\text{2}}} $

$   = \;{\text{ - 2sin}}\left( {{\text{x + }}\dfrac{{\text{h}}}{{\text{2}}}} \right){\text{sin}}\dfrac{{\text{h}}}{{\text{2}}} $

$  {{\text{f}}^\prime }({\text{x}})\; = \;\mathop {\lim }\limits_{{\text{h}} \to 0} \dfrac{{{\text{f(x + h) - f(x)}}}}{{\text{h}}} $

$   = \;\mathop {\lim }\limits_{{\text{h}} \to 0} \dfrac{{{\text{ - 2sin}}\left( {{\text{x + }}\dfrac{{\text{h}}}{{\text{2}}}} \right){\text{sin}}\dfrac{{\text{h}}}{{\text{2}}}}}{{\text{h}}} $

$   = \;\mathop {\lim }\limits_{{\text{h}} \to 0} \left[ {{\text{ - sin}}\left( {{\text{x + }}\dfrac{{\text{h}}}{{\text{2}}}} \right)} \right]\left[ {\dfrac{{{\text{sin}}\dfrac{{\text{h}}}{{\text{2}}}}}{{\dfrac{{\text{h}}}{{\text{2}}}}}} \right] $

$   = \;{\text{ - sinx}} $

अतः \[\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{cosx}}\;{\text{ = }}\;{\text{ - sinx}}\]

11. निम्नलिखित फलनों के अवकलज ज्ञात कीजिए: 

(i) \[{\text{sinx cosx}}\]

उत्तर: \[{\text{f(x)}}\] = \[{\text{sinx cosx}}\]

\[\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{(uv)}}\;{\text{ = }}\;{{\text{u}}^\prime}{\text{v + u}}{{\text{v}}^\prime}\]

$  \dfrac{{\text{d}}}{{{\text{dx}}}}{\text{(sinxcosx)}}\; = \;\left( {\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{sinx}}} \right){\text{cosx + sinx}}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{(cosx)}} $

$   = \;{\text{cosxcosx + sinx( - sinx)}} $

$   = \;{\text{co}}{{\text{s}}^{\text{2}}}{\text{x - si}}{{\text{n}}^{\text{2}}}{\text{x}} = \;{\text{cos2x}} $

(ii) \[{\text{secx}}\]

उत्तर: \[{\text{f(x)}}\] = \[{\text{secx}}\]

\[{\text{f(x + h)}}\;{\text{ = }}\;{\text{sec(x + h)}}\]

\[{\text{f(x + h) - f(x)}}\;{\text{ = }}\;{\text{sec(x + h) - secx}}\] 

= \[\dfrac{{\text{1}}}{{{\text{cos(x + h)}}}}{\text{ - }}\dfrac{{\text{1}}}{{{\text{cosx}}}}\]

$  = \;\dfrac{{{\text{cosx - cos(x + h)}}}}{{{\text{cos(x + h)cosx}}}} $

$ = \;\dfrac{{{\text{2sin}}\left( {{\text{x + }}\dfrac{{\text{h}}}{{\text{2}}}} \right){\text{sin}}\left( {\dfrac{{\text{h}}}{{\text{2}}}} \right)}}{{{\text{hcos(x + h)cosx}}}} $

$  {{\text{f}}^\prime }({\text{x}})\; = \;\mathop {\lim }\limits_{{\text{h}} \to 0} \dfrac{{{\text{f(x + h) - f(x)}}}}{{\text{h}}} $

$ = \;\mathop {\lim }\limits_{{\text{h}} \to 0} \dfrac{{{\text{sin}}\left( {{\text{x + }}\dfrac{{\text{h}}}{{\text{2}}}} \right)}}{{{\text{cos(x + h)cosx}}}}\left( {\dfrac{{{\text{sin}}\dfrac{{\text{h}}}{{\text{2}}}}}{{\dfrac{{\text{h}}}{{\text{2}}}}}} \right) $

$ = \;\dfrac{{{\text{sinx}}}}{{{\text{cosxcosx}}}} $

$ = \;\dfrac{{\text{1}}}{{{\text{cosx}}}}\dfrac{{{\text{sinx}}}}{{{\text{cosx}}}} $

$ = \;{\text{secxtanx}} $

(iii) \[{\text{5secx + 4cosx }}\]

उत्तर: \[{\text{f(x)}}\] = \[{\text{5secx + 4cosx }}\]

$  \dfrac{{\text{d}}}{{{\text{dx}}}}{\text{(secx)}}\; = \;{\text{secxtanx}} $

$  \dfrac{{\text{d}}}{{{\text{dx}}}}{\text{(cosx}})\; = \;{\text{ - sinx}} $

$  {{\text{f}}^\prime }({\text{x)}}\; = \;{\text{5}}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{(secx) + 4}}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{(cosx)}} $

$   = \;{\text{5secx tanx - 4sinx}} $

(iv) \[{\text{cosecx}}\]

उत्तर: \[{\text{f(x)}}\] = \[{\text{cosecx}}\]

\[{\text{f(x + h)}}\;{\text{ = }}\;{\text{cosec(x + h)}}\]

\[{\text{f(x + h) - f(x)}}\;{\text{ = }}\;{\text{cosec(x + h) - cosecx}}\]

$ = \;\dfrac{{\text{1}}}{{{\text{sin(x + h)}}}}{\text{ - }}\dfrac{{\text{1}}}{{{\text{sinx}}}} $

$ = \;\dfrac{{{\text{sinx - sin(x + h)}}}}{{{\text{sin(x + h)sinx}}}} $

$ = \;\dfrac{{{\text{ - 2cos}}\left( {{\text{x + }}\dfrac{{\text{h}}}{{\text{2}}}} \right){\text{sin}}\left( {\dfrac{{\text{h}}}{{\text{2}}}} \right)}}{{{\text{sin(x + h)sinx}}}} $

$  {{\text{f}}^\prime }({\text{x}})\; = \;\mathop {\lim }\limits_{{\text{h}} \to 0} \dfrac{{{\text{f(x + h) - f(x)}}}}{{\text{h}}} $

\[ = \;\mathop {\lim }\limits_{{\text{h}} \to 0} \dfrac{{{\text{ - cos}}\left( {{\text{x + }}\dfrac{{\text{h}}}{{\text{2}}}} \right)}}{{{\text{sin(x + h)sinx}}}}\left( {\dfrac{{{\text{sin}}\dfrac{{\text{h}}}{{\text{2}}}}}{{\dfrac{{\text{h}}}{{\text{2}}}}}} \right)\]

$   = \;\dfrac{{{\text{ - cosx}}}}{{{\text{sinxsinx}}}}{\text{(1)}} $

$   = \;{\text{ - cosecxcotx}} $

(v) \[{\text{3cotx + 5cosecx}}\]

उत्तर: \[{\text{f(x)}}\] = \[{\text{cotx}}\]

\[{\text{f(x + h)}}\;{\text{ = }}\;{\text{cot(x + h)}}\]

\[{\text{f(x + h) - f(x)}}\;{\text{ = }}\;{\text{cot(x + h) - cotx}}\]

$   = \;\dfrac{{{\text{cos(x + h)}}}}{{{\text{sin(x + h)}}}}{\text{ - }}\dfrac{{{\text{cosx}}}}{{{\text{sinx}}}} $

$   = \;{\text{ - }}\dfrac{{{\text{sinh}}}}{{{\text{sin(x + h)sinx}}}} $

$  {{\text{f}}^\prime }({\text{x}})\; = \;\mathop {\lim }\limits_{{\text{h}} \to 0} \dfrac{{{\text{f(x + h) - f(x)}}}}{{\text{h}}} $

$   = \;\mathop {\lim }\limits_{{\text{h}} \to 0} \dfrac{{\text{1}}}{{\text{h}}}\left[ {{\text{ - }}\dfrac{{{\text{sinh}}}}{{{\text{sin(x + h)sinx}}}}} \right] $

$   = \;{\text{ - 1}}\left( {\dfrac{{\text{1}}}{{{\text{sinxsinx}}}}} \right) $

= \[{\text{ - cose}}{{\text{c}}^{\text{2}}}{\text{x}}\]

${{\text{f}}^\prime }{\text{(x)}}\; = \;{\text{3}}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{cotx + 5}}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{cosecx}}\; $

$  {\text{ = }}\;{\text{ - 3cose}}{{\text{c}}^{\text{2}}}{\text{x - 5cosecxcotx}} $

(vi) \[{\text{5sinx - 6cosx + 7}}\]

उत्तर: \[{\text{f(x)}}\] = \[{\text{5sinx - 6cosx + 7}}\]

$  {{\text{f}}^\prime }({\text{x}})\; = \;{\text{5}}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{sinx - 6}}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{cosx + }}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{(7)}} $

 $  = \;{\text{5cosx + 6sinx}} $

(vii) \[{\text{2tanx - 7secx }}\]

उत्तर: \[{\text{f(x)}}\] = \[{\text{2tanx - 7secx }}\]

$  \dfrac{{\text{d}}}{{{\text{dx}}}}{\text{secx}}\;{\text{ = }}\;{\text{secxtanx}} $

$  \dfrac{{\text{d}}}{{{\text{dx}}}}{\text{(tanx)}}\; = \;\mathop {\lim }\limits_{{\text{h}} \to 0} \dfrac{{{\text{tan(x + h) - tanx}}}}{{\text{h}}} $

$   = \;\mathop {\lim }\limits_{{\text{h}} \to 0} \dfrac{{\text{1}}}{{\text{h}}}\left[ {\dfrac{{{\text{sin(x + h)}}}}{{{\text{cos(x + h)}}}}{\text{ - }}\dfrac{{{\text{sinx}}}}{{{\text{cosx}}}}} \right] $

$   = \;\dfrac{{\text{1}}}{{{\text{cosxcosx}}}} $

$   = \;{\text{se}}{{\text{c}}^{\text{2}}}{\text{x}} $

$  {{\text{f}}^\prime }({\text{x}})\; = \;{\text{2}}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{tanx - 7}}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{secx}} $

$   = \;{\text{2se}}{{\text{c}}^{\text{2}}}{\text{x - 7secxtanx}} $

प्रश्नावली A 1.3

1. प्रथम सिद्धांत से निम्नलिखित फलनों का अवकलन ज्ञात कीजिए (यह समझ जाए कि \[{\text{a,b,c,d,p,q,r,s}}\] निश्चित शुन्यतर अचर है \[{\text{m}}\] और \[{\text{n}}\] तथा पूर्णांक है।)

(i) \[{\text{ - x}}\]

उत्तर: माना कि \[{\text{f(x)}}\] = \[{\text{ - x}}\]

\[{\text{f(x + h)}}\;{\text{ = }}\;{\text{ - (x + h) - x - h}}\]

$  \dfrac{{{\text{d( - x)}}}}{{{\text{dx}}}}\; = \;\mathop {\lim }\limits_{{\text{h}} \to 0} \dfrac{{{\text{f(x + h) - f(x)}}}}{{\text{h}}} $

$   = \;\mathop {\lim }\limits_{{\text{h}} \to 0} \dfrac{{{\text{ - x - h + x}}}}{{\text{h}}}\; = \; - 1 $

(ii) \[{{\text{( - x)}}^{ - 1}}\]

उत्तर: माना कि \[{\text{f(x)}}\] = \[{{\text{( - x)}}^{ - 1}}\] = \[{\text{ - }}\dfrac{{\text{1}}}{{\text{x}}}\]

\[{\text{f(x + h)}}\;{\text{ = }}\;{\text{ - }}\dfrac{{\text{1}}}{{{\text{x + h}}}}\]

$  {\text{f(x + h) - f(x)}}\;{\text{ = }}\;{\text{ - }}\dfrac{{\text{1}}}{{{\text{x + h}}}}{\text{ - ( - }}\dfrac{{\text{1}}}{{\text{x}}}{\text{)}} $

$  {\text{ = }}\;\dfrac{{{\text{ - x + x + h}}}}{{{\text{(x + h)x}}}}\;{\text{ = }}\;\dfrac{{\text{h}}}{{{\text{x(x + h)}}}} $

\[\dfrac{{{\text{d(( - x}}{{\text{)}}^{{\text{ - 1}}}}{\text{)}}}}{{{\text{dx}}}}\; = \;\mathop {\lim }\limits_{{\text{h}} \to 0} \dfrac{{{\text{f(x + h) - f(x)}}}}{{\text{h}}}\; = \;\mathop {\lim }\limits_{{\text{h}} \to 0} \dfrac{{\text{h}}}{{{\text{hx(x + h)}}}}\; = \;\dfrac{{\text{1}}}{{{{\text{x}}^{\text{2}}}}}\]

(iii) \[{\text{sin(x + 1)}}\]

उत्तर: माना कि \[{\text{f(x)}}\] = \[{\text{sin(x + 1)}}\]

\[{\text{f(x + h)}}\;{\text{ = }}\;{\text{sin(x + h + 1)}}\]

$  {\text{f(x + h) - f(x)}}\;{\text{ = }}\;{\text{sin(x + h + 1) - sin(x + 1)}} $

$  {\text{ = }}\;{\text{2cos(x + 1 + }}\dfrac{{\text{h}}}{{\text{2}}}{\text{)sin(}}\dfrac{{\text{h}}}{{\text{2}}}{\text{)}} $

$  {{\text{f}}^\prime}({\text{x)}}\; = \;\mathop {\lim }\limits_{{\text{h}} \to 0} \dfrac{{{\text{f(x + h) - f(x)}}}}{{\text{h}}}\; = \;\mathop {\lim }\limits_{{\text{h}} \to 0} \dfrac{1}{{\text{h}}}2\operatorname{c} {\text{os(x + 1 + }}\dfrac{{\text{h}}}{{\text{2}}}{\text{)sin(}}\dfrac{{\text{h}}}{{\text{2}}}{\text{)}} $

$  {\text{ = }}\;\mathop {\lim }\limits_{{\text{h}} \to 0} \operatorname{c} {\text{os(x + 1 + }}\dfrac{{\text{h}}}{{\text{2}}}{\text{)sin(}}\dfrac{{\text{h}}}{{\text{2}}}{\text{)}} $

$ = \;\mathop {\lim }\limits_{{\text{h}} \to 0} \operatorname{c} {\text{os(x + 1 + }}\dfrac{{\text{h}}}{{\text{2}}}{\text{)sin(}}\dfrac{{\dfrac{{\text{h}}}{{\text{2}}}}}{{\dfrac{{\text{h}}}{{\text{2}}}}}{\text{)}}\;{\text{ = }}\;{\text{cos(x + 1)}}$

$  {\text{(}}\because \;{\text{sin(}}\dfrac{{\dfrac{{\text{h}}}{{\text{2}}}}}{{\dfrac{{\text{h}}}{{\text{2}}}}}{\text{)}}\; = \;1) $

(iv) \[{\text{cos(x - }}\dfrac{{{\pi }}}{{\text{8}}}{\text{)}}\]

उत्तर: माना कि \[{\text{f(x)}}\] = \[{\text{cos(x - }}\dfrac{{{\pi }}}{{\text{8}}}{\text{)}}\]

\[{\text{f(x + h)}}\] = \[{\text{cos(x + h - }}\dfrac{{{\pi }}}{{\text{8}}}{\text{)}}\]

$  {\text{f(x + h) - f(x)}}\;{\text{ = }}\;{\text{cos(x + h - }}\dfrac{{{\pi }}}{{\text{8}}}{\text{) - cos(x - }}\dfrac{{{\pi }}}{{\text{8}}}{\text{)}} $

$  {\text{ = }}\;{\text{ - 2sin(x - }}\dfrac{{{\pi }}}{{\text{8}}}{\text{ + }}\dfrac{{\text{h}}}{{\text{2}}}{\text{)sin}}\dfrac{{\text{h}}}{{\text{2}}} $

$  {{\text{f}}^\prime}({\text{x)}}\; = \;\mathop {\lim }\limits_{{\text{h}} \to 0} \dfrac{{{\text{f(x + h) - f(x)}}}}{{\text{h}}} $

$   = \;\mathop {\lim }\limits_{{\text{h}} \to 0} \dfrac{{{\text{ - 2sin(x - }}\dfrac{{{\pi }}}{{\text{8}}}{\text{ + }}\dfrac{{\text{h}}}{{\text{2}}}{\text{)sin}}\dfrac{{\text{h}}}{{\text{2}}}}}{{\text{h}}} $

$ = \;\mathop {\lim }\limits_{{\text{h}} \to 0} ( - \sin {\text{(x - }}\dfrac{{{\pi }}}{{\text{8}}}{\text{ + }}\dfrac{{\text{h}}}{{\text{2}}}{\text{)(sin}}\dfrac{{\dfrac{{\text{h}}}{{\text{2}}}}}{{\dfrac{{\text{h}}}{{\text{2}}}}}{\text{))}} $

$  {\text{ = }}\; - \sin {\text{(x - }}\dfrac{{{\pi }}}{{\text{8}}})\;\;\;(\because \;{\text{(sin}}\dfrac{{\dfrac{{\text{h}}}{{\text{2}}}}}{{\dfrac{{\text{h}}}{{\text{2}}}}}{\text{)}}\; = \;1) $

2. \[{\text{(x + a)}}\]

उत्तर: \[\dfrac{{{\text{d(x + a)}}}}{{{\text{dx}}}}\;{\text{ = }}\;\dfrac{{{\text{d(x)}}}}{{{\text{dx}}}}{\text{ + }}\dfrac{{{\text{d(a)}}}}{{{\text{dx}}}}\;{\text{ = }}\;{\text{1 + 0}}\;{\text{ = }}\;{\text{1}}\]

3. \[{\text{(px + q)(}}\dfrac{{\text{r}}}{{\text{x}}}{\text{ + s)}}\]

उत्तर: \[{{\text{(uv)}}^\prime}\; = \;{{\text{u}}^\prime}{\text{v + u}}{{\text{v}}^\prime}\]

\[{{\text{f}}^\prime}({\text{x)}}\; = \;\dfrac{{{\text{d(px + q)}}}}{{{\text{dx}}}}.{\text{(}}\dfrac{{\text{r}}}{{\text{x}}}{\text{ + s) + (px + q)}}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{(}}\dfrac{{\text{r}}}{{\text{x}}}{\text{ + s)}}\]

$  {\text{ = }}\;{\text{p}}{\text{.(}}\dfrac{{\text{r}}}{{\text{x}}}{\text{ + s) + (px + q)( - }}\dfrac{{\text{r}}}{{{{\text{x}}^{\text{2}}}}}{\text{)}} $

$  \dfrac{{{\text{pr}}}}{{\text{x}}}{\text{ + ps - }}\dfrac{{{\text{pr}}}}{{\text{x}}}{\text{ - }}\dfrac{{{\text{qr}}}}{{{{\text{x}}^{\text{2}}}}}\;{\text{ = }}\;{\text{ps - }}\dfrac{{{\text{qr}}}}{{{{\text{x}}^{\text{2}}}}} $

4. \[{\text{(ax + b)(cx + d}}{{\text{)}}^2}\]

उत्तर: \[\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{[(ax + b)(cx + d}}{{\text{)}}^{\text{2}}}{\text{]}}\;{\text{ = }}\;\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{[(ax + b)](cx + d}}{{\text{)}}^{\text{2}}}{\text{ + (ax + b)}}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{[(cx + d}}{{\text{)}}^{\text{2}}}{\text{]}}\]

$  {\text{ = }}\;{\text{a(cx + d}}{{\text{)}}^{\text{2}}}{\text{ + (ax + b)2c(cx + d)}} $

$  {\text{ = }}\;{\text{2c(ax + b)(cx + d) + a(cx + d}}{{\text{)}}^{\text{2}}} $

5. \[\dfrac{{{\text{(ax + b)}}}}{{{\text{(cx + d)}}}}\]

उत्तर: \[{\text{f(x)}}\] = \[\dfrac{{{\text{(ax + b)}}}}{{{\text{(cx + d)}}}}\]

$  {(\dfrac{{\text{u}}}{{\text{v}}})^\prime}\; = \;\dfrac{{{{\text{u}}^\prime}{\text{v - u}}{{\text{v}}^\prime}}}{{{{\text{v}}^{\text{2}}}}}\; $

$   = \;\dfrac{{\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{[ax + b](cx + d) - (ax + b)}}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{[cx + d]}}}}{{{{{\text{(cx + d)}}}^{\text{2}}}}} $

$ = \;\dfrac{{{\text{a(cx + d) - (ax + b)c}}}}{{{{{\text{(cx + d)}}}^{\text{2}}}}} $

$  = \;\dfrac{{{\text{acx + ad - acx - bc}}}}{{{{{\text{(cx + d)}}}^{\text{2}}}}} $

$   = \;\dfrac{{{\text{ad - bc}}}}{{{{{\text{(cx + d)}}}^{\text{2}}}}} $

6. \[\dfrac{{{\text{1 + }}\dfrac{{\text{1}}}{{\text{x}}}}}{{{\text{1 - }}\dfrac{{\text{1}}}{{\text{x}}}}}\]

उत्तर: \[{\text{f(x)}}\] = \[\dfrac{{{\text{1 + }}\dfrac{{\text{1}}}{{\text{x}}}}}{{{\text{1 - }}\dfrac{{\text{1}}}{{\text{x}}}}}\] = \[\dfrac{{{\text{x + 1}}}}{{{\text{x - 1}}}}\]

\[ = \;\dfrac{{\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{[x + 1](x - 1) - (x + 1)}}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{[x - 1]}}}}{{{{{\text{(x + 1)}}}^{\text{2}}}}}\]  \[{\text{ = }}\;\dfrac{{{\text{1(x - 1) - (x + 1)1}}}}{{{{{\text{(x - 1)}}}^{\text{2}}}}}\]  \[{\text{ = }}\;\dfrac{{{\text{ - 2}}}}{{{{{\text{(x - 1)}}}^{\text{2}}}}}\]

7. \[\dfrac{{\text{1}}}{{{\text{a}}{{\text{x}}^{\text{2}}}{\text{ + bx + c}}}}\]

उत्तर: \[{{\text{f}}^\prime}({\text{x)}} = \;\dfrac{{\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{[1](a}}{{\text{x}}^2}{\text{ + bx + c) - (1)}}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{[a}}{{\text{x}}^2}{\text{ + bx + c]}}}}{{{{{\text{(a}}{{\text{x}}^2}{\text{ + bx + c)}}}^{\text{2}}}}}\]

$ {\text{ = }}\;\dfrac{{{\text{0(a}}{{\text{x}}^{\text{2}}}{\text{ + bx + c) - (2ax + b)}}}}{{{{{\text{(a}}{{\text{x}}^{\text{2}}}{\text{ + bx + c)}}}^{\text{2}}}}} $

$  {\text{ = }}\;\dfrac{{{\text{ - (2ax + b)}}}}{{{{{\text{(a}}{{\text{x}}^{\text{2}}}{\text{ + bx + c)}}}^{\text{2}}}}} $

8. \[\dfrac{{{\text{ax + b}}}}{{{\text{p}}{{\text{x}}^{\text{2}}}{\text{ + qx + r}}}}\]

उत्तर: \[{\text{f(x)}}\] = \[\dfrac{{{\text{ax + b}}}}{{{\text{p}}{{\text{x}}^{\text{2}}}{\text{ + qx + r}}}}\]

$  {{\text{f}}^{{\prime}}}{\text{(x) = }}\;\dfrac{{\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{[ax + b](p}}{{\text{x}}^{\text{2}}}{\text{ + qx + r) - (ax + b)}}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{[p}}{{\text{x}}^{\text{2}}}{\text{ + qx + r]}}}}{{{{{\text{(p}}{{\text{x}}^{\text{2}}}{\text{ + qx + r)}}}^{\text{2}}}}}{\text{ = }}\;\dfrac{{{\text{a(p}}{{\text{x}}^{\text{2}}}{\text{ + qx + r) - (ax + b)(2px + q)}}}}{{{{{\text{(p}}{{\text{x}}^{\text{2}}}{\text{ + qx + r)}}}^{\text{2}}}}} $

$  {\text{ = }}\;\dfrac{{{\text{(ap}}{{\text{x}}^{\text{2}}}{\text{ + aqx + ar)-(2ap}}{{\text{x}}^{\text{2}}}{\text{ + (aq + 2bp)x + bq)}}}}{{{{{\text{(p}}{{\text{x}}^{\text{2}}}{\text{ + qx + r)}}}^{\text{2}}}}} $

$  {\text{ = }}\;\dfrac{{{\text{ - ap}}{{\text{x}}^{\text{2}}}{\text{ + ar - 2bpx - bq}}}}{{{{{\text{(p}}{{\text{x}}^{\text{2}}}{\text{ + qx + r)}}}^{\text{2}}}}} $

9. \[\dfrac{{{\text{p}}{{\text{x}}^{\text{2}}}{\text{ + qx + r}}}}{{{\text{ax + b}}}}\]

उत्तर: \[{\text{f(x)}}\] = \[\dfrac{{{\text{p}}{{\text{x}}^{\text{2}}}{\text{ + qx + r}}}}{{{\text{ax + b}}}}\]

$  {{\text{f}}^{{\prime}}}{\text{(x) = }}\;-\dfrac{{\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{[p}}{{\text{x}}^{\text{2}}}{\text{ + qx + r](ax + b) + (p}}{{\text{x}}^{\text{2}}}{\text{ + qx + r)}}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{[ax + b]}}}}{{{{{\text{(ax + b)}}}^{\text{2}}}}}{\text{ = }}\;\dfrac{{{\text{ - a(p}}{{\text{x}}^{\text{2}}}{\text{ + qx + r) + (ax + b)(2px + q)}}}}{{{{{\text{(ax + b)}}}^{\text{2}}}}} $

$  {\text{ = }}\;\dfrac{{{\text{2ap}}{{\text{x}}^{\text{2}}}{\text{ + 2bpx + aqx + bq - ap}}{{\text{x}}^{\text{2}}}{\text{ - aqx - ar}}}}{{{{{\text{(ax + b)}}}^{\text{2}}}}} $

$  {\text{ = }}\;\dfrac{{{\text{ap}}{{\text{x}}^{\text{2}}}{\text{ - ar + 2bpx + bq}}}}{{{{{\text{(ax + b)}}}^{\text{2}}}}} $

10. \[\dfrac{{\text{a}}}{{{{\text{x}}^{\text{4}}}}}{\text{ + }}\dfrac{{\text{b}}}{{{{\text{x}}^{\text{2}}}}}{\text{ + cosx}}\]

उत्तर: \[{\text{f(x)}}\] = \[\dfrac{{\text{a}}}{{{{\text{x}}^{\text{4}}}}}{\text{ + }}\dfrac{{\text{b}}}{{{{\text{x}}^{\text{2}}}}}{\text{ + cosx}}\]

= \[{\text{a}}{{\text{x}}^{{\text{ - 4}}}}{\text{ + b}}{{\text{x}}^{{\text{ - 2}}}}{\text{ + cosx}}\]

$  {{\text{f}}^{{\prime}}}{\text{(x)}}\;{\text{ = }}\;\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{(a}}{{\text{x}}^{{\text{ - 4}}}}{\text{) - }}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{(b}}{{\text{x}}^{{\text{ - 2}}}}{\text{) + }}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{(cosx)}} $

$  {\text{ = }}\;{\text{ - 4a}}{{\text{x}}^{{\text{ - 5}}}}{\text{ - b( - 2)}}{{\text{x}}^{{\text{ - 3}}}}{\text{ - sinx}} $

$  {\text{ = }}\;\dfrac{{{\text{4a}}}}{{{{\text{x}}^{\text{5}}}}}{\text{ + }}\dfrac{{{\text{2b}}}}{{{{\text{x}}^{\text{3}}}}}{\text{ - sinx}} $

11. \[{\text{4}}\sqrt {\text{x}} {\text{ - 2}}\]

उत्तर: \[{\text{f(x)}}\] = \[{\text{4}}\sqrt {\text{x}} {\text{ - 2}}\]

$  {{\text{f}}^{{\prime}}}{\text{(x)}}\;{\text{ = }}\;\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{(4}}\sqrt {\text{x}} {\text{) - }}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{(2)}} $

$  {\text{ = }}\;{\text{4}}{\text{.}}\dfrac{{\text{1}}}{{{\text{2}}\sqrt {\text{x}} }}{\text{ - 0}}\;{\text{ = }}\;\dfrac{{\text{2}}}{{\sqrt {\text{x}} }} $

$  {\text{(}}\because \;\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{(}}\sqrt {\text{x}} {\text{)}}\;{\text{ = }}\;\dfrac{{\text{1}}}{{{\text{2}}\sqrt {\text{x}} }}{\text{)}} $

12. \[{{\text{(ax + b)}}^{\text{n}}}\]

उत्तर: \[{\text{f(x)}}\] = \[{{\text{(ax + b)}}^{\text{n}}}\]

$ {{\text{f}}^{{\prime}}}{\text{(x)}}\;{\text{ = }}\;{\text{n(ax + b}}{{\text{)}}^{{\text{n - 1}}}}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{(ax + b)}}$

$  {\text{ = }}\;{\text{n(ax + b}}{{\text{)}}^{{\text{n - 1}}}}{\text{a}} $

13. \[{{\text{(ax + b)}}^{\text{n}}}{\text{.(cx + d}}{{\text{)}}^{\text{n}}}\]

उत्तर: \[{\text{f(x)}}\] = \[{{\text{(ax + b)}}^{\text{n}}}{\text{.(cx + d}}{{\text{)}}^{\text{n}}}\]

$  {{\text{f}}^{{\prime}}}{\text{(x) = }}\;\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{[(ax + b}}{{\text{)}}^{\text{n}}}{\text{][(cx + d}}{{\text{)}}^{\text{m}}}{\text{] + (ax + b}}{{\text{)}}^{\text{n}}}{\text{ + }}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{[(cx + d}}{{\text{)}}^{\text{m}}}{\text{]}} $

$  {\text{ = }}\;{\text{na(ax + b}}{{\text{)}}^{{\text{n - 1}}}}{{\text{(cx + d)}}^{{\text{m - 1}}}}{\text{ + (ax + b}}{{\text{)}}^{\text{n}}}{\text{mc(cx + d}}{{\text{)}}^{{\text{m - 1}}}} $

$  {\text{ = }}\;{{\text{(ax + b)}}^{{\text{n - 1}}}}{{\text{(cx + d)}}^{{\text{m - 1}}}}{\text{[na(cx + d) + mc(ax + b)]}} $

14. \[{\text{sin(x + a)}}\]

उत्तर: \[{\text{f(x)}}\] = \[{\text{sin(x + a)}}\]

माना \[{\text{x + a}}\;{\text{ = }}\;{\text{u}}\]

\[{\text{f(x)}}\] = \[{\text{sin(u)}}\]

\[{\text{x}}\] के सापेक्ष अवकलन करने पर 

$  {{\text{f}}^\prime}({\text{x)}}\;{\text{ = }}\;\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{sin(u)}}\;{\text{ = }}\;\dfrac{{\text{d}}}{{{\text{du}}}}{\text{sin(u)}}{\text{.}}\dfrac{{{\text{du}}}}{{{\text{dx}}}} $

$   = \;{\text{cos(u)}}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{(x + a)}} $

\[{\text{ = }}\;{\text{cos(x + a)}}{\text{.1}}\;{\text{ = }}\;{\text{cos(x + a)}}\]

15. \[{\text{cosecx}}{\text{.cotx}}\]

उत्तर: \[{\text{f(x)}}\] = \[{\text{cosecx}}{\text{.cotx}}\]

$  {{\text{(uv)}}^\prime}\; = \;{{\text{u}}^\prime}{\text{v + u}}{{\text{v}}^\prime} $

$  {{\text{f}}^\prime}{\text{(x)}}\;{\text{ = }}\;\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{[cosecx](cotx) + (cosecx)}}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{[cotx]}} $

$  {\text{ = }}\;{\text{( - cosecx}}{\text{.cotx)cotx + cosecx( - cose}}{{\text{c}}^{\text{2}}}{\text{x)}} $

$  {\text{ = }}\;{\text{ - cose}}{{\text{c}}^{\text{3}}}{\text{x - cosecx}}\;{\text{co}}{{\text{t}}^{\text{2}}}{\text{x}} $

16. \[\dfrac{{{\text{cosx}}}}{{{\text{1 + sinx}}}}\]

उत्तर: \[{\text{f(x)}}\] = \[\dfrac{{{\text{cosx}}}}{{{\text{1 + sinx}}}}\]

${{\text{(}}\dfrac{{\text{u}}}{{\text{v}}}{\text{)}}^\prime}\;{\text{ = }}\;\dfrac{{{{\text{u}}^\prime}{\text{v - u}}{{\text{v}}^\prime}}}{{{{\text{v}}^{\text{2}}}}}$

$  \text{f}^\prime(x) = \dfrac{{\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{[cosx](1 + sinx) - (cosx)}}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{[1 + sinx]}}}}{{{{{\text{(1 + sinx)}}}^{\text{2}}}}}{\text{ = }}\;\dfrac{{{\text{ - sinx(1 + sinx) - (cosx)(cosx)}}}}{{{{{\text{(1 + sinx)}}}^{\text{2}}}}} $

$  {\text{ = }}\;\dfrac{{{\text{ - sinx - si}}{{\text{n}}^{\text{2}}}{\text{x - co}}{{\text{s}}^{\text{2}}}{\text{x}}}}{{{{{\text{(1 + sinx)}}}^{\text{2}}}}} $

$  {\text{ = }}\;\dfrac{{{\text{ - sinx - (si}}{{\text{n}}^{\text{2}}}{\text{x + co}}{{\text{s}}^{\text{2}}}{\text{x)}}}}{{{{{\text{(1 + sinx)}}}^{\text{2}}}}} $

$  {\text{ = }}\;\dfrac{{{\text{ - sinx - 1}}}}{{{{{\text{(1 + sinx)}}}^{\text{2}}}}}$

$  {\text{ = }}\;\dfrac{{{\text{ - (sinx + 1)}}}}{{{{{\text{(1 + sinx)}}}^{\text{2}}}}} $

$  {\text{ = }}\;{\text{ - }}\dfrac{{\text{1}}}{{{\text{1 + sinx}}}} $

17. \[\dfrac{{{\text{sinx + cosx}}}}{{{\text{sinx - cosx}}}}\]

उत्तर: \[{\text{f(x)}}\] = \[\dfrac{{{\text{sinx + cosx}}}}{{{\text{sinx - cosx}}}}\]

$  {{\text{(}}\dfrac{{\text{u}}}{{\text{v}}}{\text{)}}^\prime}\;{\text{ = }}\;\dfrac{{{{\text{u}}^\prime}{\text{v - u}}{{\text{v}}^\prime}}}{{{{\text{v}}^{\text{2}}}}} $

$   \text{f}^\prime(x) = \dfrac{{\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{[sinx + cosx](sinx - cosx) - (sinx + cosx)}}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{[sinx - cosx]}}}}{{{{{\text{(sinx - cosx)}}}^{\text{2}}}}}{\text{ = }}\;\dfrac{{{{{\text{-(cosx - sinx)}}}^{\text{2}}}{\text{ - (cosx + sinx}}{{\text{)}}^{\text{2}}}}}{{{{{\text{(sinx - cosx)}}}^{\text{2}}}}} $

\[{\text{ = }}\;\dfrac{{{\text{-(si}}{{\text{n}}^{\text{2}}}{\text{x + co}}{{\text{s}}^{\text{2}}}{\text{x - 2sinx}}\;{\text{cosx) - (si}}{{\text{n}}^{\text{2}}}{\text{x + co}}{{\text{s}}^{\text{2}}}{\text{x + 2sinx}}\;{\text{cosx)}}}}{{{{{\text{(sinx - cosx)}}}^{\text{2}}}}}\]

$  {\text{ = }}\;{\text{}}\dfrac{{{\text{-1 + 2sinx}}\;{\text{cosx - 1 - 2sinx}}\;{\text{cosx}}}}{{{{{\text{(sinx - cosx)}}}^{\text{2}}}}} $

$  {\text{ = }}\;\dfrac{{{\text{ - 2}}}}{{{{{\text{(sinx - cosx)}}}^{\text{2}}}}} $

18. \[\dfrac{{{\text{secx - 1}}}}{{{\text{secx + 1}}}}\]

उत्तर: \[{\text{f(x)}}\] = \[\dfrac{{{\text{secx - 1}}}}{{{\text{secx + 1}}}}\]

$  {{\text{(}}\dfrac{{\text{u}}}{{\text{v}}}{\text{)}}^\prime}\;{\text{ = }}\;\dfrac{{{{\text{u}}^\prime}{\text{v - u}}{{\text{v}}^\prime}}}{{{{\text{v}}^{\text{2}}}}} $

$   \text{f}^\prime(x) = \dfrac{{\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{[secx - 1](secx + 1) - (secx - 1)}}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{[secx + 1]}}}}{{{{{\text{(secx + 1)}}}^{\text{2}}}}}{\text{ = }}\;\dfrac{{{\text{(secx}}\;{\text{tanx)(secx + 1) - (secx - 1)(secx}}\;{\text{tanx)}}}}{{{{{\text{(secx + 1)}}}^{\text{2}}}}} $

$  {\text{ = }}\;\dfrac{{{\text{se}}{{\text{c}}^{\text{2}}}{\text{x}}\;{\text{tanx - secx}}\;{\text{tanx - se}}{{\text{c}}^{\text{2}}}{\text{x}}\;{\text{tanx -(secx}}\;{\text{tanx)}}}}{{{{{\text{(secx + 1)}}}^{\text{2}}}}} $

$  {\text{ = }}\;{\text{ - }}\dfrac{{{\text{2secx}}\;{\text{tanx}}}}{{{{{\text{(secx + 1)}}}^{\text{2}}}}} $

19. \[{\text{si}}{{\text{n}}^{\text{n}}}{\text{x}}\]

उत्तर: \[{\text{f(x)}}\] = \[{\text{si}}{{\text{n}}^{\text{n}}}{\text{x}}\]

माना \[{\text{sinx}}\;{\text{ = }}\;{\text{u}}\]

\[{\text{f(x)}}\] = \[{{\text{u}}^{\text{n}}}\]

\[{\text{x}}\] के सापेक्ष अवकलन करने पर 

$   \text{f}^\prime(x) = \dfrac{{\text{d}}}{{{\text{dx}}}}{{\text{u}}^{\text{n}}}\;{\text{ = }}\;\dfrac{{\text{d}}}{{{\text{du}}}}{{\text{u}}^{\text{n}}}{\text{.}}\dfrac{{{\text{du}}}}{{{\text{dx}}}} $

$   = \;{\text{n}}{{\text{u}}^{{\text{n - 1}}}}\dfrac{{\text{d}}}{{{\text{dx}}}}\sin {\text{x}} $

$  {\text{ = }}\;{\text{nsi}}{{\text{n}}^{{\text{n - 1}}}}{\text{x}}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{sinx}} $

$  {\text{ = }}\;{\text{nsi}}{{\text{n}}^{{\text{n - 1}}}}{\text{x}}\;{\text{cosx}}$

$  {\text{ = }}\;{\text{ncosx}}\;{\text{si}}{{\text{n}}^{{\text{n - 1}}}}{\text{x}}$

20. \[\dfrac{{{\text{a + bsinx}}}}{{{\text{c + dcosx}}}}\]

उत्तर: \[{\text{f(x)}}\] = \[\dfrac{{{\text{a + bsinx}}}}{{{\text{c + dcosx}}}}\]

$  {{\text{(}}\dfrac{{\text{u}}}{{\text{v}}}{\text{)}}^\prime}\;{\text{ = }}\;\dfrac{{{{\text{u}}^\prime}{\text{v - u}}{{\text{v}}^\prime}}}{{{{\text{v}}^{\text{2}}}}} $

$  \text{f}^\prime(x) = \dfrac{{\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{[a + bsinx](c + dcosx) - (a + bsinx)}}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{[c + dcosx]}}}}{{{{{\text{(c + dcosx)}}}^{\text{2}}}}}{\text{ = }}\;\dfrac{{{\text{(bcosx)(c + dcosx) - (a + bsinx)( - dsinx)}}}}{{{{{\text{(c + dcosx)}}}^{\text{2}}}}} $

$  {\text{ = }}\;\dfrac{{{\text{bc}}\;{\text{cosx + ad}}\;{\text{sinx + bd(si}}{{\text{n}}^{\text{2}}}{\text{x + co}}{{\text{s}}^{\text{2}}}{\text{x)}}}}{{{{{\text{(c + dcosx)}}}^{\text{2}}}}} $

$  {\text{ = }}\;\dfrac{{{\text{bc}}\;{\text{cosx + ad}}\;{\text{sinx + bd}}}}{{{{{\text{(c + dcosx)}}}^{\text{2}}}}} $

21. \[\dfrac{{{\text{sin(x + a)}}}}{{{\text{cosx}}}}\]

उत्तर: \[{\text{f(x)}}\] = \[\dfrac{{{\text{sin(x + a)}}}}{{{\text{cosx}}}}\]

$  {{\text{(}}\dfrac{{\text{u}}}{{\text{v}}}{\text{)}}^\prime}\;{\text{ = }}\;\dfrac{{{{\text{u}}^\prime}{\text{v - u}}{{\text{v}}^\prime}}}{{{{\text{v}}^{\text{2}}}}} $

$  \text{f}^\prime(x) = \dfrac{{\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{[sin(x + a)](cosx) - (sin(x + a))}}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{[cosx]}}}}{{{{{\text{(cosx)}}}^{\text{2}}}}}{\text{ = }}\;\dfrac{{{\text{cos(x + a)(cosx) - (sin(x + a))( - sinx)}}}}{{{{{\text{(cosx)}}}^{\text{2}}}}} $

$  {\text{ = }}\;\dfrac{{{\text{cos(x + a - x)}}}}{{{{{\text{(cosx)}}}^{\text{2}}}}}$

$  {\text{ = }}\;\dfrac{{{\text{cos(a)}}}}{{{{{\text{(cosx)}}}^{\text{2}}}}} $

22. \[{{\text{x}}^{\text{4}}}{\text{(5sinx - 3cosx)}}\]

उत्तर: \[{\text{f(x)}}\] = \[{{\text{x}}^{\text{4}}}{\text{(5sinx - 3cosx)}}\]

$  {{\text{(uv)}}^\prime}\;{\text{ = }}\;{{\text{u}}^\prime}{\text{v + u}}{{\text{v}}^\prime} $

${{\text{f}^\prime(x)\text{=}}}\;\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{[}}{{\text{x}}^{\text{4}}}{\text{]((5sinx - 3cosx) + }}{{\text{x}}^{\text{4}}}{\text{}}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{[5sinx - 3cosx]}} $

$  {\text{ = }}\;{\text{4}}{{\text{x}}^{\text{3}}}{\text{(5sinx - 3cosx) + }}{{\text{x}}^{\text{4}}}{\text{(5cosx + 3sinx)}} $

$  {\text{ = }}\;{{\text{x}}^{\text{3}}}{\text{(20sinx - 12cosx + 5xcosx + 3xsinx)}} $

23. \[{\text{(}}{{\text{x}}^{\text{2}}}{\text{ + 1)cosx}}\]

उत्तर: 

f(x) = \[{\text{(}}{{\text{x}}^{\text{2}}}{\text{ + 1)cosx}}\]

$  {{\text{(uv)}}^\prime}\;{\text{ = }}\;{{\text{u}}^\prime}{\text{v + u}}{{\text{v}}^\prime} $

$   \text{f}^\prime(x) = \dfrac{{\text{d}}}{{{\text{dx}}}}{\text{[}}{{\text{x}}^2}{\text{ + 1](cosx) + (}}{{\text{x}}^2}{\text{ + 1)}}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{[cosx]}} $

\[{\text{ = }}\;{\text{2x}}\;{\text{cosx + (}}{{\text{x}}^{\text{2}}}{\text{ + 1)( -sinx)}}\]

\[{\text{ = -}}\;{{\text{x}}^{\text{2}}}{\text{sinx - sinx + 2x}}\;{\text{cosx}}\]

24. \[{\text{(a}}{{\text{x}}^{\text{2}}}{\text{ + sinx)(p + qcosx)}}\]

उत्तर: \[{\text{f(x)}}\] = \[{\text{(a}}{{\text{x}}^{\text{2}}}{\text{ + sinx)(p + qcosx)}}\]

$  {{\text{(uv)}}^\prime}\;{\text{ = }}\;{{\text{u}}^\prime}{\text{v - u}}{{\text{v}}^\prime} $]

$f^\prime(x)=\dfrac{d}{{dx}}[ax^{2}+ sinx](p + qcosx) + (ax^{2}+ sinx)\dfrac{d}{{dx}}[p + qcosx] $

$  {\text{ = }}\;{\text{(2ax + cosx)(p + qcosx) + (a}}{{\text{x}}^{\text{2}}}{\text{ + sinx)( - qsinx)}} $

25. \[{\text{(x + cosx)(x - tanx)}}\]

उत्तर: \[{\text{f(x)}}\] = \[{\text{(x + cosx)(x - tanx)}}\]

$ {{\text{(uv)}}^\prime}\;{\text{ = }}\;{{\text{u}}^\prime}{\text{v - u}}{{\text{v}}^\prime} $

$  {{\text{f}}^{{\prime}}}{\text{(x) = }}\;\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{[x + cosx](x - tanx) + (x + cosx)}}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{[x - tanx]}} $

$  {\text{ = }}\;{\text{(1 - sinx)(x - tanx) + (x + cosx)(1 - se}}{{\text{c}}^{\text{2}}}{\text{x)}} $

$  {\text{ = }}\;{\text{(1 - sinx)(x - tanx) - (x + cosx)ta}}{{\text{n}}^{\text{2}}}{\text{x}} $

26. \[\dfrac{{{\text{4x + 5sinx}}}}{{{\text{3x - 7cosx}}}}\]

उत्तर: \[{\text{f(x)}}\] = \[\dfrac{{{\text{4x + 5sinx}}}}{{{\text{3x - 7cosx}}}}\]

$  {{\text{(}}\dfrac{{\text{u}}}{{\text{v}}}{\text{)}}^\prime}\;{\text{ = }}\;\dfrac{{{{\text{u}}^\prime}{\text{v - u}}{{\text{v}}^\prime}}}{{{{\text{v}}^{\text{2}}}}}$

$  \text{f}^\prime(x) = \dfrac{{\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{[4x + 5sinx](3x + 7cosx) - (4x + 5sinx)}}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{[3x + 7cosx]}}}}{{{{{\text{(3x + 7cosx)}}}^{\text{2}}}}} $

$  {\text{ = }}\;\dfrac{{{\text{(4 + 5cosx)(3x + 7cosx) - (4x + 5sinx)(3 - 7sinx)}}}}{{{{{\text{(3x + 7cosx)}}}^{\text{2}}}}} $

$  {\text{ = }}\;\dfrac{{{\text{(12x + 28cosx + 15x}}\;{\text{cosx + 35co}}{{\text{s}}^{\text{2}}}{\text{x) - (12x - 28x}}\;{\text{sinx + 15sinx - 35si}}{{\text{n}}^{\text{2}}}{\text{x)}}}}{{{{{\text{(3x + 7cosx)}}}^{\text{2}}}}}$

$  {\text{ = }}\;\dfrac{{{\text{35 + 15x}}\;{\text{cosx + 28cosx + 28x}}\;{\text{sinx - 15sinx}}}}{{{{{\text{(3x + 7cosx)}}}^{\text{2}}}}} $

27. \[\dfrac{{{{\text{x}}^{\text{2}}}{\text{cos(}}\dfrac{{{\pi }}}{{\text{4}}}{\text{)}}}}{{{\text{sinx}}}}\]

उत्तर: \[{\text{f(x)}}\] = \[\dfrac{{{{\text{x}}^{\text{2}}}{\text{cos(}}\dfrac{{{\pi }}}{{\text{4}}}{\text{)}}}}{{{\text{sinx}}}}\]

$  {{\text{(}}\dfrac{{\text{u}}}{{\text{v}}}{\text{)}}^\prime}\;{\text{ = }}\;\dfrac{{{{\text{u}}^\prime}{\text{v - u}}{{\text{v}}^\prime}}}{{{{\text{v}}^{\text{2}}}}}$

$   \text{f}^\prime(x) = \dfrac{{\text{1}}}{{\sqrt {\text{2}} }}\dfrac{{\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{[}}{{\text{x}}^{\text{2}}}{\text{](sinx) - (}}{{\text{x}}^{\text{2}}}{\text{)}}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{[sinx]}}}}{{{{{\text{(sinx)}}}^{\text{2}}}}} $

$  {\text{ = }}\;\dfrac{{\text{1}}}{{\sqrt {\text{2}} }}\dfrac{{{\text{2x}}\;{\text{sinx - }}{{\text{x}}^{\text{2}}}{\text{cosx}}}}{{{{{\text{(sinx)}}}^{\text{2}}}}} $

28. \[\dfrac{{\text{x}}}{{{\text{1 + tanx}}}}\]

उत्तर: \[{\text{f(x)}}\] = \[\dfrac{{\text{x}}}{{{\text{1 + tanx}}}}\]

$  {{\text{(}}\dfrac{{\text{u}}}{{\text{v}}}{\text{)}}^\prime}\;{\text{ = }}\;\dfrac{{{{\text{u}}^\prime}{\text{v - u}}{{\text{v}}^\prime}}}{{{{\text{v}}^{\text{2}}}}} $

$   \text{f}^\prime(x) = \dfrac{{\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{[x](1 + tanx) - (x)}}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{[1 + tanx]}}}}{{{{{\text{(1 + tanx)}}}^{\text{2}}}}} $

$  {\text{ = }}\;\dfrac{{{\text{(1 + tanx) - x}}\;{\text{se}}{{\text{c}}^{\text{2}}}{\text{x}}}}{{{{{\text{(1 + tanx)}}}^{\text{2}}}}} $

29. \[{\text{(x + secx)(x - tanx)}}\]

उत्तर: \[{\text{f(x)}}\] = \[{\text{(x + secx)(x - tanx)}}\]

$  {{\text{(uv)}}^\prime}\;{\text{ = }}\;{{\text{u}}^\prime}{\text{v - u}}{{\text{v}}^\prime} $

$   \text{f}^\prime(x) = \dfrac{{\text{d}}}{{{\text{dx}}}}{\text{[x + secx](x - tanx) + (x + secx)}}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{[x - tanx]}} $

$  {\text{ = }}\;{\text{(1 + secx}}\;{\text{tanx)(x - tanx) + (x + secx)(1 - se}}{{\text{c}}^{\text{2}}}{\text{x)}} $

30. \[\dfrac{{\text{x}}}{{{\text{si}}{{\text{n}}^{\text{n}}}{\text{x}}}}\]

उत्तर: \[{\text{f(x)}}\] = \[\dfrac{{\text{x}}}{{{\text{si}}{{\text{n}}^{\text{n}}}{\text{x}}}}\]

$  {{\text{(}}\dfrac{{\text{u}}}{{\text{v}}}{\text{)}}^\prime}\;{\text{ = }}\;\dfrac{{{{\text{u}}^\prime}{\text{v - u}}{{\text{v}}^\prime}}}{{{{\text{v}}^{\text{2}}}}}$

$   \text{f}^\prime(x) = \dfrac{{\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{[x]( si}}{{\text{n}}^{\text{n}}}{\text{x) - (x)}}\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{[si}}{{\text{n}}^{\text{n}}}{\text{x]}}}}{{{{{\text{(si}}{{\text{n}}^{\text{n}}}{\text{x)}}}^{\text{2}}}}} $

\[{\text{ = }}\;\dfrac{{{\text{si}}{{\text{n}}^{\text{n}}}{\text{x - xn}}\;{\text{cosx}}\;{\text{si}}{{\text{n}}^{{\text{n - 1}}}}{\text{x}}}}{{{{{\text{(si}}{{\text{n}}^{\text{n}}}{\text{x)}}}^{\text{2}}}}}\]

\[{\text{ = }}\;\dfrac{{{\text{si}}{{\text{n}}^{{\text{n - 1}}}}{\text{x(sinx - nx}}\;{\text{cosx)}}}}{{{\text{(si}}{{\text{n}}^{{\text{2n}}}}{\text{x)}}}}\]

\[{\text{ = }}\;\dfrac{{{\text{sinx - nx}}\;{\text{cosx}}}}{{{\text{si}}{{\text{n}}^{{\text{n + 1}}}}{\text{x}}}}\]

NCERT Solutions for Class 11 Maths Chapter 13 Limits and Derivatives in Hindi

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