NCERT Solutions for Class 11 Maths Chapter 12 Limits and Derivatives Miscellaneous Exercise

1. Find the derivative of the following functions from first principle:

(i) -x

(ii) $\mathbf{(-x)^{-1}}$

(iii) $\mathbf{\sin (x+1)}$

(iv) $\mathbf{\cos \left(x-\dfrac{\pi}{8}\right)}$

Ans: (i) Let $f(x)=-x$. Accordingly, $f(x+h)=-(x+h)$

By first principle, $f^{\prime}(x)=\lim _{n \rightarrow 0} \dfrac{f(x+h)-f(x)}{h}$

$=\lim _{h \rightarrow 0} \dfrac{-(x+h)-(-x)}{h}$

$=\lim _{n \rightarrow 0} \dfrac{-x-h+x}{h}$

$=\lim _{n \rightarrow 0} \dfrac{-h}{h}$

$=\lim _{h \rightarrow 0}(-1)=-1$

(ii) Let $f(x)=(-x)^{-1}=\dfrac{1}{-x}=\dfrac{-1}{x} .$ Accordingly, $f(x+h)=\dfrac{-1}{(x+h)}$

By first principle,

$f^{\prime}(x)=\lim _{n \rightarrow 0} \dfrac{f(x+h)-f(x)}{h}$

$=\lim _{h \rightarrow 0} \dfrac{1}{h}\left[\dfrac{-1}{(x+h)}-\left(\dfrac{-1}{x}\right)\right]$

$=\lim _{n \rightarrow 0} \dfrac{1}{h}\left[\dfrac{-x+(x+h)}{x(x+h)}\right]$

$=\lim _{h \rightarrow 0} \dfrac{1}{h}\left[\dfrac{h}{x(x+h)}\right]$

$=\lim _{n \rightarrow 0} \dfrac{1}{x(x+h)}$

$=\dfrac{1}{x \cdot X}=\dfrac{1}{x^{2}}$

(iii) Let $f(x)=\sin (x+1)$. Accordingly, $f(x+h)=\sin (x+h+1)$

By first principle,

$f^{\prime}(x)=\lim _{n \rightarrow 0} \dfrac{f(x+h)-f(x)}{h}$

$=\lim _{n \rightarrow 0} \dfrac{1}{h}[\sin (x+h+1)-\sin (x+1)]$

$=\lim _{n \rightarrow 0} \dfrac{1}{h}\left[2 \cos \left(\dfrac{x+h+1+x+1}{2}\right) \sin \left(\dfrac{x+h+1-x-1}{2}\right)\right]$

$=\lim _{n \rightarrow 0} \dfrac{1}{h}\left[2 \cos \left(\dfrac{2 x+h+2}{2}\right) \sin \left(\dfrac{h}{2}\right)\right]$

$=\lim _{n \rightarrow 0}\left[\cos \left(\dfrac{2 x+h+2}{2}\right) \cdot \dfrac{\sin \left(\dfrac{h}{2}\right)}{\left(\dfrac{h}{2}\right)}\right]$

$\begin{array}{l}=\lim _{n \rightarrow 0} \dfrac{1}{h} \cos \left(\dfrac{2 x+h+2}{2}\right) \cdot \lim _{\dfrac{b}{2} \rightarrow} \dfrac{\sin \left(\dfrac{h}{2}\right)}{h} \dfrac{h}{\left.\dfrac{h}{2}\right)} \quad\left[\text { As } h \rightarrow 0 \Rightarrow \dfrac{h}{2} \rightarrow 0\right]\\=\cos \left(\dfrac{2 x+0+2}{2}\right) \cdot 1 \quad\left[\lim _{h \rightarrow 0} \dfrac{\sin x}{x}=1\right]\\=\cos (x+1)\\\text (iv) Let f(x)=\cos \left(x-\dfrac{\pi}{8}\right) Accordingly, f(x+h)-\cos \left(x+h-\dfrac{\pi}{8}\right)\\\text { By first principle, }\\f^{\prime}(x)=\lim _{n \rightarrow 0} \dfrac{f(x+f)-f(x)}{h}\\=\lim _{h \rightarrow 0} \dfrac{1}{h}\left[\cos \left(x+h-\dfrac{\pi}{8}\right)-\cos \left(x-\dfrac{\pi}{8}\right)\right]\\=\lim _{n \rightarrow 0} \dfrac{1}{h}\left[-2 \sin \dfrac{\left(x+h-\dfrac{\pi}{8}+x-\dfrac{\pi}{8}\right)}{2} \sin \left(\dfrac{x+h-\dfrac{\pi}{8}-x+\dfrac{\pi}{8}}{2}\right)\right]\\=\lim _{n \rightarrow 0} \dfrac{1}{h}\left[-2 \sin \left(\dfrac{2 x+h-\dfrac{\pi}{4}}{2}\right) \sin \left(\dfrac{h}{2}\right)\right]\\=\lim _{n \rightarrow 0}\left[-\sin \left(\dfrac{2 x+h-\dfrac{\pi}{4}}{2}\right) \dfrac{\sin \left(\dfrac{h}{2}\right)}{\left(\dfrac{h}{2}\right)}\right]\\\left.=\lim _{n \rightarrow 0}\left[-\sin \left(\dfrac{2 x+h-\dfrac{\pi}{4}}{2}\right)\right] \cdot \lim _{\dfrac{\pi}{2} \rightarrow 0} \dfrac{\sin \left(\dfrac{h}{2}\right)}{\left(\dfrac{h}{2}\right)} \quad \text { [As } h \rightarrow 0 \Rightarrow \dfrac{h}{2} \rightarrow 0\right]\\=-\sin \left(\dfrac{2 x+0-\dfrac{\pi}{4}}{2}\right) \cdot 1\end{array}$

$=-\sin \left(x-\dfrac{\pi}{8}\right)$

2. Find the derivative of the following functions (it is to be understood that a, b, c, d. p, q, r and s are fixed non-zero constants and $m$ and $n$ are integers): $(x+a)$

Ans: Let $f(x)=x+a$. Accordingly. $f(x+h)-x+h+a$ By first principle,

$f^{\prime}(x)=\lim _{n \rightarrow 0} \dfrac{f(x+h)-f(x)}{h}$

$\lim _{n \rightarrow 0} \dfrac{x+h+a-x-a}{h}$

$\lim _{n \rightarrow 0}\left(\dfrac{h}{h}\right)$

$-\lim _{n \rightarrow 0}(1)$

$=1$

3. Find the derivative of the following functions (it is to be understood that $a, b, c$. d, $p, q, r$ and s are fixed non- zero constants and $m$ and $n$ are integers): $\mathbf{(p x+q)\left(\dfrac{r}{x}+s\right)}$

Ans: Let $f(x)=(p x+q)\left(\dfrac{r}{x}+s\right)$

By Leibnitz product rule.

$f^{\prime}(x)=(p x+q)\left(\dfrac{r}{x}+s\right)^{\prime}+\left(\dfrac{r}{x}+s\right)(p x+q)^{\prime}$

$-(p x+q)\left(r x^{-1}+s\right)^{\prime}+\left(\dfrac{r}{x}+s\right)(p)$

$-(p x+q)\left(-n x^{2}\right)+\left(\dfrac{r}{x}+s\right) p$

$=(p x+q)\left(\dfrac{-r}{x^{2}}\right)+\left(\dfrac{r}{x}+s\right) p$

$=\dfrac{-p x}{x}-\dfrac{q r}{x^{2}}+\dfrac{p r}{x}+p s$

$p s-\dfrac{q r}{x^{2}}$

4. Find the derivative of the following functions (it is to be understood that $a, b, c$, d, $p, q, r$ and $s$ are fixed non-zero constants and $m$ and $n$ are integers): $\mathbf{(ax+b)(c x+d)^{2}}$

Ans: Let $f^{\prime}(x)=(a x+b)(c x+d)^{2}$

By Leibnitz product rule,

$f^{\prime}(x)=(a x+b) \dfrac{d}{d x}(c x+d)^{2} \dfrac{d}{d x}(a x+b)$

$(a x+b) \dfrac{d}{d x}\left(c^{2} x^{2}+2 c d x^{2}\right)+(c x+d)^{2} \dfrac{d}{d x}(a x+b)$

$(a x+b)\left[\dfrac{d}{d x}\left(c^{2} x^{2}\right)+\dfrac{d}{d x}(2 c d x)+\dfrac{d}{d x} d^{2}\right]+(c x+d)^{2}\left[\dfrac{d}{d x} a x+\dfrac{d}{d x} b\right]$

$=(a x+b)\left(2 c^{2} x+2 c d\right)+(c x+d)^{2} a$

$-2 c(a x+b)(c x+d)+a(c x+d)^{2}$

5. Find the derivative of the following functions (it is to be understood that $a, b, c$, d, $p, q, r$ and $s$ are fixed non zero constants and $m$ and $n$ are integers): $\mathbf{\dfrac{a x+b}{c x+d}}$

Ans: Let $f(x)=\dfrac{a x+b}{c x+d}$

By quotient rule,

$f(x)=\dfrac{(c x+d) \dfrac{d}{d x}(a x+b)-(a x+b) \dfrac{d}{d x}(c x+d)}{(c x+d)^{2}}$

$=\dfrac{(c x+d)(a)-(a x+d)(c)}{(c x+d)^{2}}$

$\dfrac{a c x+a d-a c x-b c}{(c x+d)^{2}}$

$\dfrac{a d-b c}{(c x+d)^{2}}$

6. Find the derivative of the following functions (it is to be understood that $a, b, c$, d, $p, q, r$ and $s$ are fixed non-zero constants and $m$ and $n$ are integers): $\mathbf{\dfrac{1+\dfrac{1}{x}}{1-\dfrac{1}{x}}}$

Ans: Let$f(x)=\dfrac{1+\dfrac{1}{x}}{1-\dfrac{1}{x}}=\dfrac{\dfrac{x+1}{x}}{\dfrac{x-1}{x}}=\dfrac{x+1}{x-1}$, where $x \neq 0$

By quotient rule, $f^{\prime}(x)=\dfrac{(x-1) \dfrac{d}{d x}(x-1)-(x+1) \dfrac{d}{d x}(x-1)}{(x-1)^{2}}, x \neq 0,1$

$=\dfrac{(x-1)(1)-(x+1)(1)}{(x-1)^{2}}, x \neq 0,1$

$\dfrac{x-1-x-1}{(x-1)^{2}}, x \neq 0,1$

$\dfrac{-2}{(x-1)^{2}}, x \neq 0,1$

7. Find the derivative of the following functions (it is to be understood that $a, b, c$, d, $p, q, r$ and s are fixed non-zero constants and $m$ and $n$ are integers) $\mathbf{: \dfrac{1}{a x^{2}+b x+c}}$

Ans: Let $f(x)=\dfrac{1}{a x^{2}+b x+c}$

By quotient rule,

$f^{\prime}(x)=\dfrac{\left(a x^{2}+b x+c\right) \dfrac{d}{d x}(1)-\dfrac{d}{d x}\left(a x^{2}+b x+c\right)}{\left(a x^{2}+b x+c\right)^{2}}$

$\dfrac{\left(a x^{2}+b x+c\right)(0)-(2 a x+b)}{\left(a x^{2}+b x+c\right)^{2}}$

$\dfrac{-(2 a x+b)}{\left(a x^{2}+b x+c\right)^{2}}$

8. Find the derivative of the following functions (it is to be understood that $a, b, c$ d, p, q, r and s are fixed non-zero constants and m and $n$ are integers): $\mathbf{\dfrac{ax+b}{p x^{2}+q x+r}}$

Ans: Let $f(x)=\dfrac{a x+b}{p x^{2}+q x+r}$

By quotient rule,

$f^{\prime}(x)=\dfrac{\left(p x^{2}+q x+r\right) \dfrac{d}{d x}(a x+b)-(a x+b) \dfrac{d}{d x}\left(p x^{2}+q x+r\right)}{\left(p x^{2}+q x+r\right)^{2}}$

$=\dfrac{\left(p x^{2}+q x+r\right)(a)-(a x+b)(2 p x+q)}{\left(p x^{2}+q x+r\right)^{2}}$

$=\dfrac{a p x^{2}+a q x+a r-a q x+2 n p x+b q}{\left(p x^{2}+q x+r\right)^{2}}$

$\dfrac{-a p x^{2}+2 b p x+a r-b q}{\left(p x^{2}+q x+r\right)^{2}}$

9. Find the derivative of the following functions (it is to be understood that $a, b, c$. d, $p$, q, $r$ and s are fixed non-zero constants and $m$ and $n$ are integers): $\mathbf{\dfrac{p x^{2}+q x+r}{ax+b}}$

Ans: Let $f(x)=\dfrac{p x^{2}+q x+r}{a x+b}$

By the quotient rule,

$\dot{f}(x)=\dfrac{(a x+b) \dfrac{d}{d x}\left(p x^{2}+q x+\eta\right)-\left(p x^{2}+q x+r\right) \dfrac{d}{d x}(a x+b)}{(a x+b)^{2}}$

$\dfrac{(a x+b)(2 p x+q)-\left(p x^{2}+q x+r\right)(a)}{(a x+b)^{2}}$

$=\dfrac{2 a p x^{2}+a q x+2 b p x+b q-a q x^{2}-a q x-a r}{(a x+b)^{2}}$

$\dfrac{a p x^{2}+2 b p x+b q-a r}{(a x+b)^{2}}$

10. Find the derivative of the following functions (it is to be understood that $a, b, c$, d, $p$, q, $r$ and $s$ are fixed non-zero constants and $m$ and $n$ are integers): $\mathbf{\dfrac{a}{x^{4}}-\dfrac{b}{x^{2}}+\cos x}$

Ans: Let $f(x)=\dfrac{a}{x^{4}}-\dfrac{b}{x^{2}}+\cos x$

$f^{\prime}(x)=\dfrac{d}{d x}\left(\dfrac{a}{x^{4}}\right)-\dfrac{d}{d x}\left(\dfrac{a}{x^{2}}\right)+\dfrac{d}{d x}(\cos x)$

$a \dfrac{d}{d x}\left(x^{-4}\right)-b \dfrac{d}{d x}\left(x^{2}\right)+\dfrac{d}{d x}(\cos x)$

$-a\left(-4 x^{-5}\right)-b\left(-2 x^{3}\right)+(-\sin x) \quad\left[\dfrac{d}{d x}\left(x^{n}\right)=n x^{n-1}\right.$ and $\left.\dfrac{d}{d x}(\cos x)=-\sin x\right]$

$\dfrac{-4 a}{x^{5}}+\dfrac{2 b}{x^{3}}-\sin x$

11. Find the derivative of the following functions (it is to be understood that $a, b, c$, d, $p, q, r$ and $s$ are fixed nonzero constants and $m$ and $n$ are integers): $\mathbf{4 \sqrt{x}-2}$

Ans: Let $f(x)=4 \sqrt{x}-2$

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

$=4 \dfrac{d}{d x}\left(x^{\dfrac{1}{2}}\right)-0=4\left(\dfrac{1}{2} x^{\dfrac{1}{2}}\right)$

$=\left(2 x^{-\dfrac{1}{2}}\right)=\dfrac{2}{\sqrt{x}}$

12. Find the derivative of the following functions (it is to be understood that $a, b, c$, d, $p, q, r$ and $s$ are fixed non-zero constants and $m$ and $n$ are integers): $\mathbf{(ax+b)^{n}}$

Ans: Let $f(x)=(a x+b)^{n} .$ Accordingly, $f(x+h)-\{a(x+h)+b\}^{n}-(a x+a h+b)^{n}$

By first principle,

$f^{\prime}(x)=\lim _{n \rightarrow 0} \dfrac{f(x+h)-f(x)}{h}$

$=\lim _{h \rightarrow 0} \dfrac{(a x+a h+b)-(a x+b)^{n}}{h}$

$=\lim _{h \rightarrow 0} \dfrac{(a x+b)^{n}\left(1+\dfrac{a h}{a x+b}\right)^{n}-(a x+b)^{n}}{h}$

$=(a x+b)^{n} \lim _{n \rightarrow 0} \dfrac{1}{h}\left[\left\{1+n\left(\dfrac{a h}{a x+b}\right)+\dfrac{n(n-1)}{2}\left(\dfrac{a h}{a x+b}\right)^{2}+\cdots\right\}-1\right] \quad$ (using binomial theorem)

$=(a x+b)^{n} \lim _{n \rightarrow 0} \dfrac{1}{h}\left[\left(\dfrac{a h}{a x+b}\right)+\dfrac{n(n-1) a^{2} h^{2}}{2(a x+b)^{2}}+\cdots\right.$ (Terms containing higher degrees of $\left.\left.h\right)\right]$

$=(a x+b)^{n} \lim _{n \rightarrow 0}\left[\dfrac{n a}{(a x+b)}+\dfrac{n(n-1) \nexists^{7} h^{2}}{2(a x+b)^{2}}+\cdots\right]$

$=(a x+b)^{n}\left[\dfrac{n a}{(a x+b)}+0\right]$

$=n a \dfrac{(a x+b)^{n}}{a x+b}$

$-n a(a x+b)^{n-1}$

13. Find the derivative of the following functions (it is to be understood that $a, b, c$, d, $p, q, r$ and $s$ are fixed non-zero constants and $m$ and $n$ are integers): $\mathbf{(a x+b)^{n}(c x+d)^{m}}$

Ans: Let $f(x)=(a x+b)^{n}(c x+d)^{m}$

By Leibnitz product rule,

$f^{\prime}(x)=(a x+b)^{n} \dfrac{d}{d x}(c x+d)^{m}+(c x+d)^{m} \dfrac{d}{d x}(a x+b)^{n}$

Now let $f_{1}(x)=(c x+d)^{m}$

$f_{1}(x+h)=(c x+c h+d)^{m}$

$f_{1}^{\prime}(x)=\lim _{n \rightarrow 0} \dfrac{f_{1}(x+h)-f_{1}(x)}{h}$

$=\lim _{n \rightarrow 0} \dfrac{(c x+c h+d)^{m}-(c x+d)^{m}}{h}$

$=(c x+d)^{m} \lim _{n \rightarrow 0} \dfrac{1}{h}\left[\left(1+\dfrac{c h}{c x+d}\right)^{m}-1\right]$

$=(c x+d)^{m} \lim _{h \rightarrow 0} \dfrac{1}{h}\left[\left(1+\dfrac{m c h}{(c x+d)}+\dfrac{m(m-1)}{2} \dfrac{c^{2} h^{2}}{(c x+d)^{2}}+\cdots\right)^{m}-1\right]$

$=(c x+d)^{m} \lim _{n \rightarrow 0} \dfrac{1}{h}\left[\dfrac{m c h}{(c x+d)}+\dfrac{m(m-1) c^{2} h^{2}}{2(c x+d)^{2}}+\cdots\right.$ (Terms containing higher degree oh $\left.\left.h\right)\right]$

$=(c x+d)^{m} \lim _{h \rightarrow 0}\left[\dfrac{m c}{(c x+d)}+\dfrac{m(m-1) c^{2} h^{2}}{2(c x+d)^{2}}+\cdots\right]$

$=(C x+a)^{m}\left[\dfrac{m c h}{(c x+d)}+0\right]$

$=\dfrac{m c(c x+d)^{m}}{(c x+d)}$

$=m c(c x+d)^{m-1}$

$\dfrac{d}{d x}(c x+d)^{m}=m d(x+d)^{m-1}$

Similarly, $\dfrac{d}{d x}(a x+b)^{n}=n a(a x+b)^{n-1}$... (3)

Therefore, from (1), (2), and (3), we obtain

$f^{\prime}(x)=(a x+b)^{n}\left\{m c(c x+d)^{m-1}\right\}+(c+d)^{m}\left\{n a(a x+b)^{n-1}\right\}$

$=(a x+b)^{n-1}(c x+d)^{m-1}[m c(a x+b)+n a(c x+d)]$

14. Find the derivative of the following functions (it is to be understood that $a, b, c$. d, $p, q, r$ and s are fixed non-zero constants and $m$ and $n$ are integers): $\mathbf{\sin (x+a)}$

Ans: Let, $f(x)=\sin (x+a)$

$f(x+h)=\sin (x+h+a)$

By first principle, $f^{\prime}(x)=\lim _{n \rightarrow 0} \dfrac{f(x+h)-f(x)}{h}$

$=\lim _{n \rightarrow 0} \dfrac{\sin (x+h+a)-\sin (x+a)}{h}$

$=\lim _{n \rightarrow 0} \dfrac{1}{h}\left[2 \cos \left(\dfrac{x+h+a+x+a}{2}\right) \sin \left(\dfrac{x+h+a-x-a}{2}\right)\right]$

$=\lim _{n \rightarrow 0} \dfrac{1}{h}\left[2 \cos \left(\dfrac{2 x+2 a+h}{2}\right) \sin \left(\dfrac{h}{2}\right)\right]$

$=\lim _{h \rightarrow 0}\left[\cos \left(\dfrac{2 x+2 a+h}{2}\right)\left[\dfrac{\sin \left(\dfrac{h}{2}\right)}{\left(\dfrac{h}{2}\right)}\right]\right]$

$=\lim _{h \rightarrow 0} \cos \left(\dfrac{2 x+2 a+h}{2}\right) \cdot \lim _{\dfrac{h}{2} \rightarrow 0}\left[\dfrac{\sin \left(\dfrac{h}{2}\right)}{\left(\dfrac{h}{2}\right)}\right] \quad\left[\right.$ As $\left.h \rightarrow 0 \Rightarrow \dfrac{h}{2} \rightarrow 0\right]$

$=\cos \left(\dfrac{2 x+2 a}{2}\right) \times 1 \quad\left[\lim _{n \rightarrow 0} \dfrac{\sin x}{x}=1\right]$

$=\cos (x+a)$

15. Find the derivative of the following functions (it is to be understood that $a, b, c$, d, $p, q, r$ and $s$ are fixed non-zero constants and $m$ and $n$ are integers): $\mathbf{\operatorname{cosec} x \cot x}$

Ans: Let $f(x)=\operatorname{cosec} x \cot x$

By Leibnitz product rule,

$f^{\prime}(x)=\operatorname{cosec} x(\cot x)^{\prime}+\cot x(\operatorname{cosec} x)^{\prime} \ldots .(1)$

Let $f_{1}(x)=\cot x .$ Accordingly, $f_{1}(x+h)=\cot (x+h)$

By first principle,

$f^{\prime \prime}(x)=\lim _{n \rightarrow 0} \dfrac{f(x+h)-f(x)}{h}$

$=\lim _{h \rightarrow 0} \dfrac{\cot (x+h)-\cot (x)}{h}$

$=\lim _{h \rightarrow 0} \dfrac{1}{h}\left(\dfrac{\cos (x+h)}{\sin (x+h)}-\dfrac{\cos (x)}{\sin x}\right)$

$=\lim _{n \rightarrow 0} \dfrac{1}{h}\left(\dfrac{\sin x \cos (x+h)-\cos x \sin (x+h)}{\sin x \sin (x+h)}\right)$

$=\lim _{h \rightarrow 0} \dfrac{1}{h}\left(\dfrac{\sin (x-x+h)}{\sin x \sin (x+h)}\right)$

$=\dfrac{1}{\sin x^{n \rightarrow 0}} \lim _{h} \dfrac{1}{h}\left[\dfrac{\sin (-h)}{\sin (x+h)}\right]$

$=\dfrac{-1}{\sin x}\left(\lim _{n \rightarrow 0} \dfrac{\sin h}{h}\right)\left(\lim _{n \rightarrow 0} \dfrac{1}{\sin (x+h)}\right)$

$=\dfrac{-1}{\sin x} \cdot 1 \cdot\left(\lim _{n \rightarrow 0} \dfrac{1}{\sin (x+0)}\right)$

$=\dfrac{-1}{\sin ^{2} x}$

$=-\operatorname{cosec}^{2} x$

$\therefore(\cot x)^{\prime}=-\operatorname{cosec}^{2} x \quad \ldots$ (2)

Now, let $f_{2}(x)=\operatorname{cosec} x .$ Accordingly, $f_{2}(x+h)=\operatorname{cosec}(x+h)$

By first principle, $f_{2}(x)=\lim _{n \rightarrow 0} \dfrac{f_{2}(x+h)-f_{2}(x)}{h}$

$=\lim _{h \rightarrow 0} \dfrac{1}{h}[\operatorname{cosec}(x+h)-\operatorname{cosec}(x)]$

$=\lim _{n \rightarrow 0} \dfrac{1}{h}\left(\dfrac{1}{\sin (x+h)}-\dfrac{1}{\sin x}\right)$

$=\lim _{n \rightarrow 0} \dfrac{1}{h}\left(\dfrac{\sin x-\sin (x+h)}{\sin x \sin (x+h)}\right)$

$=\dfrac{1}{\sin x} \cdot \lim _{h \rightarrow 0} \dfrac{1}{h}\left[\dfrac{2 \cos \left(\dfrac{x+x+h}{2}\right) \sin \left(\dfrac{x-x-h}{2}\right)}{\sin (x+h)}\right]$

$=\dfrac{1}{\sin x} \cdot \lim _{n \rightarrow 0} \dfrac{1}{h}\left[\dfrac{2 \cos \left(\dfrac{2 x+h}{2}\right) \sin \left(\dfrac{-h}{2}\right)}{\sin (x+h)}\right]$

$=\dfrac{1}{\sin x} \cdot \lim _{h \rightarrow 0}\left[\dfrac{-\sin \left(\dfrac{h}{2}\right)}{\left(\dfrac{h}{2}\right)} \cdot \dfrac{\cos \left(\dfrac{2 x+h}{2}\right)}{\sin (x+h)}\right]$

$=\dfrac{-1}{\sin x} \cdot \lim _{h \rightarrow 0} \dfrac{\sin \left(\dfrac{h}{2}\right)}{\left(\dfrac{h}{2}\right)} \cdot \lim _{n \rightarrow 0} \dfrac{\cos \left(\dfrac{2 x+h}{2}\right)}{\sin (x+h)}$

$=\dfrac{-1}{\sin x} \cdot 1 \cdot \dfrac{\cos \left(\dfrac{2 x+h}{2}\right)}{\sin (x+0)}$

$=\dfrac{-1}{\sin x} \cdot \dfrac{\cos x}{\sin x}$

$=-\operatorname{cosec} x \cdot \cot x$

$\therefore(\operatorname{cosec} x)^{\prime}=-\operatorname{cosec} x \cdot \cot x$

From (1), (2), and (3), we obtain

$f^{\prime}(x)=\operatorname{cosec} x\left(-\operatorname{cosec}^{2} x\right)+\cot x(-\operatorname{cosec} x \cot x)$

$=-\operatorname{cosec}^{3} x-\cot ^{2} x \operatorname{cosec} x$

16. Find the derivative of the following functions (it is to be understood that $a, b, c$, d, $p, q, r$ and $s$ are fixed non-zero constants and $m$ and $n$ are integers): $\mathbf{\dfrac{\cos x}{1+\sin x}}$

Ans: Let $f(x)=\dfrac{\cos x}{1+\sin x}$

By the quotient rule,

$f^{\prime}(x)=\dfrac{(1+\sin x) \dfrac{d}{d x}(\cos x)-(\cos x) \dfrac{d}{d x}(1+\sin x)}{(1+\sin x)^{2}}$

$=\dfrac{(1+\sin x)(-\sin x)-(\cos x)(\cos x)}{(1+\sin x)^{2}}$

$=\dfrac{-\sin x-\sin ^{2} x-\cos ^{2} x}{(1+\sin x)^{2}}$

$=\dfrac{-\sin x-\left(\sin ^{2} x+\cos ^{2} x\right)}{(1+\sin x)^{2}}$

$=\dfrac{-\sin x-1}{(1+\sin x)^{2}}$

$=\dfrac{-(1-\sin x)}{(1+\sin x)^{2}}$

$=\dfrac{-1}{(1+\sin x)^{2}}$

17. Find the derivative of the following functions (it is to be understood that $a, b, c$, d, $p, q, r$ and $s$ are fixed non-zero constants and m and n are integers): $\mathbf{\dfrac{\sin x+\cos x}{\sin x-\cos x}}$

Ans:17: Let $f(x)=\dfrac{\sin x+\cos x}{\sin x-\cos x}$

By the quotient rule,

$f^{\prime \prime}(x)=\dfrac{(\sin x-\cos x) \dfrac{d}{d x}(\sin x+\cos x)-(\sin x+\cos x) \dfrac{d}{d x}(\sin x-\cos x)}{(\sin x+\cos x)^{2}}$

$=\dfrac{(\sin x-\cos x)(\cos x-\sin x)-(\sin x+\cos x)(\cos x+\sin x)}{(\sin x+\cos x)^{2}}$

$=\dfrac{-(\sin x-\cos x)^{2}-(\sin x+\cos x)^{2}}{(\sin x+\cos x)^{2}}$

$=\dfrac{-\left[\sin ^{2} x+\cos ^{2} x-2 \sin x \cos x+\sin ^{2} x+\cos ^{2} x+2 \sin x \cos x\right]}{(\sin x+\cos x)^{2}}$

$=\dfrac{-[1+1]}{(\sin x-\cos x)^{2}}$

$=\dfrac{-2}{(\sin x-\cos x)^{2}}$

18. Find the derivative of the following functions (it is to be understood that $a, b, c$, d. $p, q, r$ and $s$ are fixed non-zero constants and $m$ and $n$ are integers): $\mathbf{\dfrac{\sec x-1}{\sec x+1}}$

Ans: Let $f(x)=\dfrac{\sec x-1}{\sec x+1}$

$f(x)=\dfrac{\dfrac{1}{\cos x}-1}{\dfrac{1}{\cos x}+1}=\dfrac{1-\cos x}{1+\cos x}$

By the quotient rule,

$f^{\prime}(x)=\dfrac{(1+\cos x) \dfrac{d}{d x}(1-\cos x)-(1-\cos x) \dfrac{d}{d x}(1+\cos x)}{(1+\cos x)^{2}}$

$=\dfrac{(1+\cos x)(\sin x)-(1-\cos x)(-\sin x)}{(1+\cos x)^{2}}$

$=\dfrac{\sin x+\cos x \sin x+\sin x-\sin x \cos x}{(1+\cos x)^{2}}$

$=\dfrac{2 \sin x}{(1+\cos x)^{2}}$

$=\dfrac{2 \sin x}{\left(1+\dfrac{1}{\sec x}\right)^{2}}=\dfrac{2 \sin x}{\dfrac{(\sec x+1)^{2}}{\sec ^{2} x}}$

$=\dfrac{2 \sin x \sec ^{2} x}{(\sec x+1)^{2}}$

$=\dfrac{\dfrac{2 \sin x}{\cos x} \sec x}{(\sec x+1)^{2}}$

$=\dfrac{2 \sec x \tan x}{(\sec x+1)^{2}}$

19. Find the derivative of the following functions (it is to be understood that $a, b, c$, d. $p, q, r$ and s are fixed non-zero constants and $m$ and $n$ are integers): $\mathbf{\sin ^{n} x}$

Ans: Let $y=\sin ^{n} x$

Accordingly, for $n=1, y=\sin x$

$\therefore \dfrac{d y}{d x}=\cos x$, i.e., $\dfrac{d}{d x} \sin x=\cos x$

For $n=2, y=\sin ^{2} x$.

$\therefore \dfrac{d y}{d x}=\dfrac{d}{d x}(\sin x \sin x)$

$=(\sin x)^{\prime}\left(\sin x+\sin x(\sin x)^{\prime} \quad\right.$ (By Leibnitz product rule)

$=\cos x \sin x+\sin x \cos x$

$=2 \sin x \cos x$

$\ldots .(1)$

For $n=3, y=\sin ^{3} x$

$\therefore \dfrac{d y}{d x}=\dfrac{d}{d x}\left(\sin x \sin ^{2} x\right)$

$=(\sin x)^{\prime} \sin ^{2} x+\sin x(\sin x)^{\prime}$

(By Leibnitz product rule)

$-\cos x \sin ^{2} x+\sin x(2 \sin x \cos x) \quad[$ Using $(1)]$

$=\cos x \sin ^{2} x+\sin ^{2} x \cos x$

$=3 \sin ^{2} x \cos x$

We assert that $\dfrac{d}{d x}\left(\sin ^{n} x\right)=n \sin ^{(n-1)} x \cos x$

Let our assertion be true for $n=k$.

i.e., $\dfrac{d}{d x}\left(\sin ^{k} x\right)=k \sin ^{(k-1)} x \cos x \quad \ldots .$ (2)

Consider,

$\dfrac{d}{d x}\left(\sin ^{k+1} x\right)=\dfrac{d}{d x}\left(\sin x \sin ^{(k)} x\right)$

$=(\sin x)^{\prime} \sin ^{k} x+\sin x\left(\sin ^{k} x\right)^{n}$

(By Leibnitz product rule)

$=\cos x \sin ^{k} x+\sin x\left(k \sin ^{k-1} \cos x\right) \quad[$ Using $(2)]$

$=\cos x \sin ^{k} x+2 \sin ^{k} x \cos x$

$-(k+1) \sin ^{k} x \cos x$

Thus, our assertion is true for $n=k+1$.

Hence, by mathematical induction, $\dfrac{d}{d x}\left(\sin ^{n} x\right)=n \sin ^{(n-1)} x \cos x$

20. Find the derivative of the following functions (it is to be understood that $a, b, c$, d, $p, q, r$ and s are fixed non-zero constants and $m$ and n are integers): $\mathbf{\dfrac{a+b \sin x}{c+d \cos x}}$

Ans: Let $f(x)=\dfrac{a+b \sin x}{c+d \cos x}$

By the quotient rule,

$f^{\prime}(x)=\dfrac{(c+d \cos x) \dfrac{d}{d x}(a+b \sin x)-(a+b \sin x) \dfrac{d}{d x}(c+d \cos x)}{(c+d \cos x)^{2}}$

$=\dfrac{(c+d \cos x)(b \cos x)-(a+b \sin x)(-d \sin x)}{(c+d \cos x)^{2}}$

$=\dfrac{c b \cos x+b d \cos ^{2} x+a d \sin x+b d \sin ^{2} x}{(c+d \cos x)^{2}}$

$=\dfrac{b c \cos x+a d \sin x+b d\left(\cos ^{2} x+\sin ^{2} x\right)}{(C+d \cos x)^{2}}$

$=\dfrac{b c \cos x+a d \sin x+b d}{(c+d \cos x)^{2}}$

21. Find the derivative of the following functions (it is to be understood that $a, b, c$, d, $p, q, r$ and s are fixed non-zero constants and m and $n$ are integers): $\mathbf{\dfrac{\sin (x+a)}{\cos x}}$

Ans: Let $f(x)=\dfrac{\sin (x+a)}{\cos x}$

By the quotient rule,

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

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

Let $g(x)-\sin (x+a) .$ Accordingly,$g(x+h)=\sin (x+h+a)$

By first principle,

$g^{\prime}(x)=\lim _{h \rightarrow 0} \dfrac{g(x+h)-g(x)}{h}$

$=\lim _{n \rightarrow 0} \dfrac{1}{h}[\sin (x+h+a)-\sin (x+a)]$

$=\lim _{n \rightarrow 0} \dfrac{1}{h}\left[2 \cos \left(\dfrac{x+h+a+x+a}{2}\right) \sin \left(\dfrac{x+h+a-x-a}{2}\right)\right]$

$=\lim _{n \rightarrow 0} \dfrac{1}{h}\left[2 \cos \left(\dfrac{2 x+2 a+h}{2}\right) \sin \left(\dfrac{h}{2}\right)\right]$

$=\lim _{n \rightarrow 0}\left[\cos \left(\dfrac{2 x+2 a+h}{h}\right)\left\{\dfrac{\sin \left(\dfrac{h}{2}\right)}{\left(\dfrac{h}{2}\right)}\right\}\right]$

$=\lim _{n \rightarrow 0} \cos \left(\dfrac{2 x+2 a+h}{h}\right) \cdot \lim _{n \rightarrow 0}\left\{\dfrac{\sin \left(\dfrac{h}{2}\right)}{\left(\dfrac{h}{2}\right)}\right\} \quad\left[\right.$ As $\left.h \rightarrow 0 \Rightarrow \dfrac{h}{2} \rightarrow 0\right]$

$=\left(\cos \dfrac{2 x+2 a}{2}\right) \times 1 \quad\left[\lim _{n \rightarrow 0} \dfrac{\sin h}{h}=1\right]$

$=\cos (x+a) \quad \ldots$ (ii)

From (i) and (ii), we obtain $f^{\prime}(x)=\dfrac{\cos x \cos (x+a)+\sin x \sin (x+a)}{\cos ^{2} x}$

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

$=\dfrac{\cos a}{\cos ^{2} x}$

22. Find the derivative of the following functions (it is to be understood that $a, b, c$, d. $p, q, r$ and $s$ are fixed non-zero constants and $m$ and $n$ are integers) $\mathbf{: x^{4}(5 \sin x-3 \cos x)}$

Ans: Let $f(x)=x^{4}(5 \sin x-3 \cos x)$

By product rule.

$f^{\prime}(x)=x^{4} \dfrac{d}{d x}(5 \sin x-3 \cos x)+(5 \sin x-3 \cos x) \dfrac{d}{d x}\left(x^{4}\right)$

$=x^{4}\left[5 \dfrac{d}{d x}(\sin x)-3 \dfrac{d}{d x}(\cos x)\right]+(5 \sin x-3 \cos x) \dfrac{d}{d x}\left(x^{4}\right)$

$=x^{4}[5 \cos x-3(-\sin x)]+(5 \sin x-3 \cos x)\left(4 x^{3}\right)$

$=x^{3}[5 x \cos x+3 x \sin x+20 \sin x-12 \cos x]$

23. Find the derivative of the following functions (it is to be understood that $a, b, c$, d, p, q, r and s are fixed non-zero constants and m and n are integers): $\mathbf{\left(x^{2}+1\right) \cos x}$

Ans: Let $f(x)=\left(x^{2}+1\right) \cos x$

By product rule.

$f^{\prime}(x)=\left(x^{2}+1\right) \dfrac{d}{d x}(\cos x)+\cos x \dfrac{d}{d x}\left(x^{2}+1\right)$

$=\left(x^{2}+1\right)(-\sin x)+\cos x(2 x)$

$=-x^{2} \sin x-\sin x+2 x \cos x$

24. Find the derivative of the following functions (it is to be understood that $a, b, c$, d, $p, q, r$ and s are fixed non-zero constants and $m$ and $n$ are integers): $\mathbf{\left(ax^{2}+\sin x\right)(p+q \cos x)}$

Ans: Let $f(x)=\left(a x^{2}+\sin x\right)(p+q \cos x)$

By product rule.

$f^{\prime}(x)=\left(a x^{2}+\sin x\right) \dfrac{d}{d x}(p+q \cos x)+(p+q \cos x) \dfrac{d}{d x}\left(a x^{2}+\sin x\right)$

$=\left(a x^{2}+\sin x\right)(-q \sin x)+(p+q \cos x)(2 a x+\cos x)$

$=-q \sin x\left(a x^{2}+\sin x\right)+(p+q \cos x)(2 a x+\cos x)$

25. Find the derivative of the following functions (it is to be understood that $a, b, c$, d. $p, q, r$ and $s$ are fixed non-zero constants and $m$ and $n$ are integers): $\mathbf{(x+\cos x)(x-\tan x)}$

Ans: Let $f(x)=(x+\cos x)(x-\tan x)$

By product rule,

$f^{\prime}(x)=(x+\cos x) \dfrac{d}{d x}(x-\tan x)+(x-\tan x) \dfrac{d}{d x}(x+\cos x)$

$=(x+\cos x)\left[\dfrac{d}{d x}(x)-\dfrac{d}{d x}(\tan x)\right]+(x-\tan x)(1-\sin x)$

$=(x+\cos x)\left[1-\dfrac{d}{d x}(\tan x)\right]+(x-\tan x)(1-\sin x)$

Let $g(x)=\tan x .$ Accordingly,$g(x+h)=\tan (x+h)$

By first principle,

$g^{\prime \prime}(x)=\lim _{n \rightarrow 0} \dfrac{g(x+h)-g(x)}{h}$

$=\lim _{h \rightarrow 0} \dfrac{\tan (x+h)-\tan (x)}{h}$

$=\lim _{n \rightarrow 0} \dfrac{1}{h}\left[\dfrac{\sin (x+h)}{\cos (x+h)}-\dfrac{\sin x}{\cos x}\right]$

$=\lim _{n \rightarrow 0} \dfrac{1}{h}\left[\dfrac{\sin (x+h) \cos x-\sin x \cos (x+h)}{\cos x \cos (x+h)}\right]$

$=\dfrac{1}{\cos x^{n \rightarrow 0}} \lim _{h} \dfrac{1}{h}\left[\dfrac{\sin (x+h-x)}{\cos (x+h)}\right]$

$=\dfrac{1}{\cos x^{h \rightarrow 0}} \lim _{h} \dfrac{1}{h}\left[\dfrac{\sin h}{\cos (x+h)}\right]$

$=\dfrac{1}{\cos x}\left(\lim _{n \rightarrow 0} \dfrac{\sin h}{h}\right)\left(\lim _{n \rightarrow 0} \dfrac{1}{\cos (x+h)}\right)$

$=\dfrac{1}{\cos x} \cdot \cdot\left(\dfrac{1}{\cos (x+0)}\right)$

$=\dfrac{1}{\cos ^{2} x}$

$=\sec ^{2} x \quad$... (ii)

Therefore, from (i) and (ii). We obtain

$f^{\prime}(x)=(x+\cos x)\left(1-\sec ^{2} x\right)+(x-\tan x)(1-\sin x)$

$=(x+\cos x)\left(-\tan ^{2} x\right)+(x-\tan x)(1-\sin x)$

$=-\tan ^{2} x(x+\cos x)+(x-\tan x)(1-\sin x)$

26. Find the derivative of the following functions (it is to be understood that $a, b, c$, d, $p, q, r$ and s are fixed non-zero constants and m and $n$ are integers): $\mathbf{\dfrac{4 x+5 \sin x}{3 x+7 \cos x}}$

Ans: Let $f(x)=\dfrac{4 x+5 \sin x}{3 x+7 \cos x}$

Quotient rule,

$f^{\prime}(x)=\dfrac{(3 x+7 \cos x) \dfrac{d}{d x}(4 x+5 \sin x)-(4 x+5 \sin x) \dfrac{d}{d x}(3 x+7 \cos x)}{(3 x+7 \cos x)^{2}}$

$=\dfrac{(3 x+7 \cos x)\left[4 \dfrac{d}{d x}(x)+5 \dfrac{d}{d x}(\sin x)\right]-(4 x+5 \sin x)\left[3 \dfrac{d}{d x}(x)+7 \dfrac{d}{d x}(\cos x)\right]}{(3 x+7 \cos x)^{2}}$

$=\dfrac{(3 x+7 \cos x)[4 x+5 \cos x]-(4 x+5 \sin x)[3-7 \sin x]}{(3 x+7 \cos x)^{2}}$

$=\dfrac{12 x+15 x \cos x+28 x \cos x+35 \cos ^{2} x-12 x+28 x \sin x-15 \sin x+35\left(\cos ^{2} x+\sin ^{2} x\right)}{(3 x+7 \cos x)^{2}}$

$=\dfrac{15 x \cos x+28 \cos x+28 x \sin x-15 \sin x+35\left(\cos ^{2} x+\sin ^{2} x\right)}{(3 x+7 \cos x)^{2}}$

$=\dfrac{35+15 x \cos x+28 \cos x+28 x \sin x-15 \sin x}{(3 x+7 \cos x)^{2}}$

27. Find the derivative of the following functions (it is to be understood that $a, b, c$, d, $p, q, r$ and $s$ are fixed non-zero constants and $m$ and $n$ are integers): $\mathbf{\dfrac{x^{2} \cos \left(\dfrac{\pi}{4}\right)}{\sin x}}$

Ans: Let $\mathrm{f}(\mathrm{x})=\dfrac{x^{2} \cos \left(\dfrac{\pi}{4}\right)}{\sin x}$

By quotient rule, $f^{\prime}(x)=\cos \left(\dfrac{\pi}{4}\right)\left[\dfrac{\sin x \dfrac{d}{d x}\left(x^{2}\right)-x^{2} \dfrac{d}{d x}(\sin x)}{\sin ^{2} x}\right]$

$=\cos \left(\dfrac{\pi}{4}\right)\left[\dfrac{\sin x(2 x)-x^{2}(\cos x)}{\sin ^{2} x}\right]$

$=\dfrac{x \cos \dfrac{\pi}{4}[2 \sin x-x \cos x]}{\sin ^{2} x}$

28. Find the derivative of the following functions (it is to be understood that $a, b, c$. d, p, q, r and s are fixed non-zero constants and $m$ and $n$ are integers): $\mathbf{\dfrac{x}{1+\tan x}}$

Ans: Let $f(x)=\dfrac{x}{1+\tan x}$

$f(x)=\dfrac{(1+\tan x) \dfrac{d}{d x}(x)-(x) \dfrac{d}{d x}(1+\tan x)}{(1+\tan x)^{2}}$

$f^{\prime}(x)=\dfrac{(1+\tan x)-x \dfrac{d}{d x}(1+\tan x)}{(1+\tan x)^{2}}$

Let $g(x)=1+\tan x .$ Accordingly $g(x+h)=1+\tan (x+h)$.

By first principle, $\dot{g}(x)=\lim _{n \rightarrow 0} \dfrac{g(x+h)-g(x)}{h}$

$\lim _{h \rightarrow 0}\left[\dfrac{1+\tan (x+h)-1-\tan (x)}{h}\right]$

$\lim _{n \rightarrow 0} \dfrac{1}{h}\left[\dfrac{\sin (x+h)}{\cos (x+h)}-\dfrac{\sin x}{\cos x}\right]$

$\lim _{n \rightarrow 0} \dfrac{1}{h}\left[\dfrac{\sin (x+h) \cos x-\sin x \cos x(x+h)}{\cos (x+h) \cos x}\right]$

$\lim _{n \rightarrow 0} \dfrac{1}{h}\left[\dfrac{\sin (x+h-x)}{\cos (x+h) \cos x}\right]$

$\lim _{h \rightarrow 0} \dfrac{1}{h}\left[\dfrac{\sinh }{\cos (x+h) \cos x}\right]$

$-\left(\lim _{n \rightarrow 0} \dfrac{\sinh }{h}\right) \cdot\left(\lim _{n \rightarrow 0} \dfrac{1}{\cos (x+h) \cos x}\right)$

$-1 \times \dfrac{1}{\cos ^{2}}=\sec ^{2} x$

$\Rightarrow \dfrac{d}{d x}\left(1+\tan ^{2} x\right)=\sec ^{2} x$

From (i) and (ii), we obtain

$\dot{f}(x)=\dfrac{1+\tan x-x \sec ^{2} x}{(1+\tan x)^{2}}$

29. Find the derivative of the following functions (it is to be understood that $\mathrm{a}, \mathrm{b}, \mathrm{c}$, d, p, q, r and s are fixed non-zero constants and m and n are integers): $\mathbf{(x+\sec x)(x-\tan x)}$

Ans: Let $f(x)=(x+\sec x)(x-\tan x)$

By product rule.

$f(x)=(x+\sec x) \dfrac{d}{d x}(x-\tan x)+(x-\tan x) \dfrac{d}{d x}(x+\sec x)$

$-(x+\sec x)\left[\dfrac{d}{d x}(x)-\dfrac{d}{d x} \tan x\right]+(x-\tan x)\left[\dfrac{d}{d x}(x)-\dfrac{d}{d x} \sec x\right]$

$\left.-f(x+\sec x)\left[1-\dfrac{d}{d x} \tan x\right)\right]+(x-\tan x)\left[1+\dfrac{d}{d x} \sec x\right]$

$\ldots(\mathrm{i})$

Let $f_{1}(x)=\tan x, f_{2}(x)=\sec x$

Accordingly, $f_{1}(x+h) \cdot \tan (x+h)$ and $f_{2}(x+h)-\sec (x+h)$

$f_{1}^{\prime}(x)=\lim _{n \rightarrow 0}\left(\dfrac{f_{1}(x+h)-f_{1}(x)}{h}\right)$

$=\lim _{h \rightarrow 0}\left[\dfrac{\tan (x+h)-\tan (x)}{h}\right]$

$\begin{array}{l}-\lim _{n \rightarrow 0} \dfrac{1}{h}\left[\dfrac{\sin (x+h)}{\cos (x+h)}-\dfrac{\sin x}{\cos x}\right]\\-\lim _{n \rightarrow 0} \dfrac{1}{h}\left[\dfrac{\sin (x+h) \cos x-\sin x \cos x(x+h)}{\cos (x+h) \cos x}\right]\\-\lim _{n \rightarrow 0} \dfrac{1}{h}\left[\dfrac{\sin (x+h-x)}{\cos (x+h) \cos x}\right]\\-\lim _{n \rightarrow 0} \dfrac{1}{h}\left[\dfrac{\sinh }{\cos (x+h) \cos x}\right]\\-\left(\lim _{n \rightarrow 0} \dfrac{\sinh }{h}\right) \cdot\left(\lim _{n \rightarrow 0} \dfrac{1}{\cos (x+h) \cos x}\right)\\-1 \times \dfrac{1}{\cos ^{2}}=\sec ^{2} x\\\Rightarrow \dfrac{d}{d x}\left(1+\tan ^{2} x\right)=\sec ^{2} x\\f_{2}^{\prime}(x)=\lim _{n \rightarrow 0}\left(\dfrac{f_{2}+(x+h)-f_{2}(x)}{h}\right)\\=\lim _{h \rightarrow 0}\left(\dfrac{\sec (x+h)-\sec (x)}{h}\right)\\=\lim _{n \rightarrow 0} \dfrac{1}{h}\left(\dfrac{1}{\cos (x+h)}-\dfrac{1}{\cos x}\right)\\=\lim _{n \rightarrow 0} \dfrac{1}{h}\left(\dfrac{\cos x-\cos (x+h)}{\cos (x+h) \cos x}\right)\\=\dfrac{1}{\cos x^{\prime \rightarrow 0}} \lim _{h} \dfrac{1}{h}\left[\dfrac{-2 \sin \left(\dfrac{x+x+h}{2}\right) \cdot \sin \left(\dfrac{x-x-h}{2}\right)}{\cos (x+h)}\right]\\=\dfrac{1}{\cos x} \lim _{h \rightarrow 0} \dfrac{1}{h}\left[\dfrac{-2 \sin \left(\dfrac{2 x+h}{2}\right) \cdot \sin \left(\dfrac{-h}{2}\right)}{\cos (x+h)}\right]\end{array}$

$=\dfrac{1}{\cos x} \lim _{h \rightarrow 0} \dfrac{1}{h}\left[\dfrac{\sin \left(\dfrac{2 x+h}{2}\right)\left\{\dfrac{\sin\left(\dfrac{h}{2}\right)}{\left(\dfrac{h}{2}\right)}\right\}}{\cos (x+h)}\right]$

$=\sec x \dfrac{\left\{\lim _{n \rightarrow 0} \sin \left(\dfrac{2 x+h}{2}\right)\right\}\left\{\lim _{\dfrac{h}{2} \rightarrow 0} \dfrac{\sin \left(\dfrac{h}{2}\right)}{\left(\dfrac{h}{2}\right)}\right\}}{\lim _{n \rightarrow 0} \cos (x+h)}$

$=\sec x \cdot \dfrac{\sin x \cdot 1}{\cos x}$

$\Rightarrow \dfrac{d}{d x} \sec x=\sec x \tan x$

From (i). (ii), and (iii), we obtain

$f^{\prime}(x)=(x+\sec x)\left(1-\sec ^{2} x\right)+(x-\tan x)(1+\sec x \tan x)$

30. Find the derivative of the following functions (it is to be understood that $a, b, c$, d, $p, q, r$ and s are fixed non-zero constants and $m$ and $n$ are integers): $\mathbf{\dfrac{x}{\sin ^{n} x}}$

Ans: Let $\mathrm{f}(\mathrm{x})=\dfrac{x}{\sin ^{n} x}$

By quotient rule, $f^{\prime}(x)=\dfrac{\sin ^{n} x \dfrac{d}{d x} x-x \dfrac{d}{d x} \sin ^{n} x}{\sin ^{2 n} x}$

It can be easily shown that $\dfrac{d}{d x} \sin ^{n} x=n \sin ^{n-1} x \cos x$

Therefore,

$f^{\prime}(x)=\dfrac{\sin ^{n} \times \dfrac{d}{d x} x-x \dfrac{d}{d x} \sin ^{n} x}{\sin ^{2 n} x}$

$=\dfrac{\sin ^{n} x \cdot 1-x\left(\operatorname{nin}^{n-1} x \cos x\right)}{\sin ^{2 n} x}$

$=\dfrac{\sin ^{n-1} x(\sin x-n x \cos x)}{\sin ^{2 n} x}$

$=\dfrac{\sin x-n x \cos x}{\sin ^{n+1} x}$

Conclusion

NCERT Solutions for Class 11 Maths Chapter 12, focused on Limits and Derivatives, are fundamental for understanding the basics of calculus. This chapter focuses on finding out the rates of change (derivatives) of functions and how they solve particular values (limits). To fully understand these concepts, you must focus on solving various kinds of problems. The basic foundation for advanced mathematical studies and practical applications is a knowledge of limits and derivatives. Using NCERT Solutions regularly provides complete exam preparation and improves your mathematical problem-solving skills.

Class 11 Maths Chapter 12: Exercises Breakdown

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Chapter-Specific NCERT Solutions for Class 11 Maths

Given below are the chapter-wise NCERT Solutions for Class 11 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

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