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# NCERT Solutions for Class 12 Maths Miscellaneous Exercise Chapter 5- Continuity and Differentiability

Last updated date: 02nd Aug 2024
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## NCERT for Class 12 Maths Chapter 5 Miscellaneous Exercise Solutions - Free PDF Download

Class 12 Maths Chapter 5 miscellaneous exercise solutions include solutions to all Miscellaneous Exercise problems. Miscellaneous Chapter 5 Class 12 is based on the ideas presented in Maths Chapter 5. To excel in the board exam, download the NCERT Solutions for Class 12 Maths Continuity and Differentiability Miscellaneous Exercise in PDF format and practice them offline. Students can download the revised Class 12 Maths NCERT Solutions from our page, which is prepared so that you can understand it easily.

Table of Content
1. NCERT for Class 12 Maths Chapter 5 Miscellaneous Exercise Solutions - Free PDF Download
2. Access NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability
2.1Miscellaneous Exercise
3. Class 12 Maths Chapter 5: Exercises Breakdown
4. CBSE Class 12 Maths Chapter 5 Other Study Materials
5. Chapter-Specific NCERT Solutions for Class 12 Maths
FAQs

Class 12 Chapter 5 Maths Miscellaneous Exercise Solutions are aligned with the updated CBSE guidelines for Class 12, ensuring students are well-prepared for exams. Access the Class 12 Maths Syllabus here.

Competitive Exams after 12th Science

## Access NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability

### Miscellaneous Exercise

Differentiate w.r.t. to  $\mathbf{x}$ , the following function.

1.$\mathbf{y=(3}{{\mathbf{x}}^{\mathbf{2}}}\mathbf{-9x+5}{{\mathbf{)}}^{\mathbf{9}}}$ .

Ans: The given function is

$\text{y=(3}{{\text{x}}^{\text{2}}}\text{-9x+5}{{\text{)}}^{\text{9}}}$ .

Differentiating both sides with respect to  $\text{x}$  gives

\begin{align}&\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{d}}{\text{dx}}{{\text{(3}{{\text{x}}^{\text{2}}}\text{-9x+5)}}^{\text{9}}}\\&\text{=9(3}{{\text{x}}^{\text{2}}}\text{-9x+5}{{\text{)}}^{\text{8}}}\text{}\!\!\times\!\!\text{}\dfrac{\text{d}}{\text{dx}}\text{(3}{{\text{x}}^{\text{2}}}\text{-9x+5)}\\&\text{=9(3}{{\text{x}}^{\text{2}}}\text{-9x+5}{{\text{)}}^{\text{8}}}\text{}\!\!\times\!\!\text{(6x-9x)} \\ & \text{=9(3}{{\text{x}}^{\text{2}}}\text{-9x+5}{{\text{)}}^{\text{8}}}\text{ }\!\!\times\!\!\text{3(2x-3)}\\&\text{=27(3}{{\text{x}}^{\text{2}}}\text{-9x+5}{{\text{)}}^{\text{8}}}\text{(2x-3)} \\ \end{align}

2.$\mathbf{y=si}{{\mathbf{n}}^{\mathbf{3}}}\mathbf{x+co}{{\mathbf{s}}^{\mathbf{6}}}\mathbf{x}$ .

Ans: The given function is

$\text{y=si}{{\text{n}}^{\text{3}}}\text{x+co}{{\text{s}}^{\text{6}}}\text{x}$ .

Differentiating both sides with respect to  $\text{x}$  gives

\begin{align}&\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{d}}{\text{dx}}\text{=(si}{{\text{n}}^{\text{3}}}\text{x)+}\dfrac{\text{d}}{\text{dx}}\text{(co}{{\text{s}}^{\text{6}}}\text{x)}\\&\text{=3si}{{\text{n}}^{\text{2}}}\text{x}\!\!\times\!\!\text{}\dfrac{\text{d}}{\text{dx}}\text{(sinx)+6co}{{\text{s}}^{\text{5}}}\text{x}\dfrac{\text{d}}{\text{dx}}\text{(cosx)}\\&\text{=3si}{{\text{n}}^{\text{2}}}\text{x}\!\!\times\!\!\text{cosx+6co}{{\text{s}}^{\text{5}}}\text{x(-sinx)}\\&\text{=3sin}^2\text{xcosx(sinx-2co}{{\text{s}}^{\text{4}}}\text{x)} \\ \end{align}

3.  $\mathbf{y=(5x}{{\mathbf{)}}^{\mathbf{3cos2x}}}$ .

Ans: The given function is  $\text{y=(5x}{{\text{)}}^{\text{3cos2x}}}$ .

First, take the logarithm of both sides of the function.

$\log y=3\cos 2x\log 5x$ .

Then, differentiating both sides with respect to  $\text{x}$  gives

\begin{align}&\dfrac{\text{1}}{\text{y}}\dfrac{\text{dy}}{\text{dx}}\text{=3}\left[\text{log5}\text{.}\dfrac{\text{d}}{\text{dx}}\text{(cos2x)+cos2x}\text{.}\dfrac{\text{d}}{\text{dx}}\text{(log5x)}\right]\\&\Rightarrow\dfrac{\text{dy}}{\text{dx}}\text{=3y}\left[\text{log5x(-sin2x)}\text{.}\dfrac{\text{d}}{\text{dx}}\text{(2x)+cos2x}\text{.}\dfrac{\text{1}}{\text{5x}}\text{.}\dfrac{\text{d}}{\text{dx}}\text{(5x)} \right] \\ & \Rightarrow \dfrac{\text{dy}}{\text{dx}}\text{=3y}\left[ \text{-2sin2xlog5x+}\dfrac{\text{cos2x}}{\text{x}} \right] \\ & \Rightarrow \dfrac{\text{dy}}{\text{dx}}\text{=3y}\left[\dfrac{\text{3cos2x}}{\text{x}}\text{-6sin2xlog5x} \right] \\ \end{align}

Hence,  $\dfrac{\text{dy}}{\text{dx}}\text{=(5x}{{\text{)}}^{\text{3cos2x}}}\left[ \dfrac{\text{3cos2x}}{\text{x}}\text{-6sin2xlog5x} \right]$ .

4. $\text{y=si}{{\text{n}}^{\text{-1}}}\left( \text{x}\sqrt{\text{x}} \right)\text{,}\,\,\,\,\text{0}\le \text{x}\le 1$ .

Ans: The given function is  $\text{y=si}{{\text{n}}^{\text{-1}}}\left( \text{x}\sqrt{\text{x}} \right)$ .

Then, differentiating both sides with respect to  $\text{x}$  by using the chain rule gives

\begin{align}&\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{d}}{\text{dx}}\text{si}{{\text{n}}^{\text{-1}}}\left(\text{x}\sqrt{\text{x}}\right)\\&=\dfrac{\text{1}}{\sqrt{\text{1-}{{\left(\text{x}\sqrt{\text{x}}\right)}^{2}}}}\text{}\!\!\times\!\!\text{}\dfrac{\text{d}}{\text{dx}}\left(\text{x}\sqrt{\text{x}}\right)\\&=\dfrac{\text{1}}{\sqrt{\text{1-}{{\text{x}}^{\text{3}}}}}\text{.}\dfrac{\text{d}}{\text{dx}}\left({{\text{x}}^{\dfrac{3}{2}}}\right)\\&=\dfrac{\text{1}}{\sqrt{\text{1-}{{\text{x}}^{\text{3}}}}}\text{}\!\!\times\!\!\text{}\dfrac{\text{3}}{\text{2}}\text{.}{{\text{x}}^{\dfrac{\text{1}}{\text{2}}}}\\&=\dfrac{\text{3}\sqrt{\text{x}}}{\text{2}\sqrt{\text{1-}{{\text{x}}^{\text{3}}}}} \\ \end{align}

Hence,

$\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{3}}{\text{2}}\sqrt{\dfrac{\text{x}}{\text{1-}{{\text{x}}^{\text{3}}}}}$ .

5. $\text{y=}\dfrac{\text{co}{{\text{s}}^{\text{-1}}}\dfrac{\text{x}}{\text{2}}}{\sqrt{\text{2x+7}}}\text{,}\,\,\,\,\,\text{-2}$<$x$<$2$ .

Ans: The given function is

$\text{y=}\dfrac{\text{co}{{\text{s}}^{\text{-1}}}\dfrac{\text{x}}{\text{2}}}{\sqrt{\text{2x+7}}}$ .

Then, differentiating both sides with respect to  $\text{x}$  using the quotient rule gives

\begin{align}&\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\sqrt{\text{2x+7}}\dfrac{\text{d}}{\text{dx}}\left(\text{co}{{\text{s}}^{\text{-1}}}\dfrac{\text{x}}{\text{2}} \right)\text{-}\left( \text{co}{{\text{s}}^{\text{-1}}}\dfrac{\text{x}}{\text{2}} \right)\dfrac{\text{d}}{\text{dx}}\left(\sqrt{\text{2x+7}}\right)}{{{\left(\sqrt{\text{2x+7}}\right)}^{\text{2}}}}\\&=\dfrac{\sqrt{\text{2x+7}}\left[\dfrac{\text{-1}}{\sqrt{\text{1-}{{\left(\dfrac{\text{x}}{\text{2}}\right)}^{\text{2}}}}}\text{.}\dfrac{\text{d}}{\text{dx}}\left( \dfrac{\text{x}}{\text{2}} \right) \right]\text{-}\left( \text{co}{{\text{s}}^{\text{-1}}}\dfrac{\text{x}}{\text{2}}\right)\dfrac{\text{1}}{\text{2}\sqrt{\text{2x+7}}}\text{.}\dfrac{\text{d}}{\text{dx}}\text{(2x+7)}}{\text{2x+7}}\\&=\dfrac{\sqrt{\text{2x+7}}\dfrac{\text{-1}}{\sqrt{\text{4-}{{\text{x}}^{\text{2}}}}}\text{-}\left( \text{co}{{\text{s}}^{\text{-1}}}\dfrac{\text{x}}{\text{2}} \right)\dfrac{\text{2}}{\text{2}\sqrt{\text{2x+7}}}}{\text{2x+7}}\\&=\dfrac{\text{-}\sqrt{\text{2x+7}}}{\sqrt{\text{4-}{{\text{x}}^{\text{2}}}}\text{}\!\!\times\!\!\text{(2x+7)}}\text{-}\dfrac{\text{co}{{\text{s}}^{\text{-1}}}\dfrac{\text{x}}{\text{2}}}{\left( \sqrt{\text{2x+7}} \right)\left( \text{2x+7} \right)} \\ \end{align}

Hence,  $\dfrac{\text{dy}}{\text{dx}}\text{=-}\left[ \dfrac{\text{1}}{\sqrt{4-x^2}\sqrt{2x+7}}\text{+}\dfrac{\text{co}{{\text{s}}^{\text{-1}}}\dfrac{\text{x}}{\text{2}}}{{{\left( \text{2x+7} \right)}^{\dfrac{\text{3}}{\text{2}}}}} \right]$ .

6. $\mathbf{y=co}{{\mathbf{t}}^{\mathbf{-1}}}\left[ \dfrac{\sqrt{\mathbf{1+sinx}}\mathbf{+}\sqrt{\mathbf{1-sinx}}}{\sqrt{\mathbf{1+sinx}}\mathbf{-}\sqrt{\mathbf{1-sinx}}} \right]\mathbf{,}\text{ }\mathbf{0}<\mathbf{x}<\mathbf{2}$ .

Ans: The given function is  $\text{y=co}{{\text{t}}^{\text{-1}}}\left[ \dfrac{\sqrt{\text{1+sinx}}\text{+}\sqrt{\text{1-sinx}}}{\sqrt{\text{1+sinx}}\text{-}\sqrt{\text{1-sinx}}} \right]$                 ……. (1)

Now,

$\dfrac{\sqrt{\text{1+sinx}}\text{+}\sqrt{\text{1-sinx}}}{\sqrt{\text{1+sinx}}\text{-}\sqrt{\text{1-sinx}}}$

\begin{align}&=\dfrac{\left(\sqrt{\text{1+sinx}}\text{+}\sqrt{\text{1-sinx}}\right)}{\left(\sqrt{\text{1+sinx}}\text{-}\sqrt{\text{1-sinx}}\right)\sqrt{\text{1+sinx}}\text{+}\sqrt{\text{1-sinx}}}\\&=\dfrac{\text{(1+sinx)+(1-sinx)+2}\sqrt{\text{(1+sinx)-(1-sinx)}}}{\text{(1+sinx)-(1-sinx)}}\\&=\dfrac{\text{2+2}\sqrt{\text{1-si}{{\text{n}}^{\text{2}}}\text{x}}}{\text{2sinx}}\\&=\dfrac{\text{1+cosx}}{\text{sinx}}\\&=\dfrac{\text{2co}{{\text{s}}^{\text{2}}}\dfrac{\text{x}}{\text{2}}}{\text{2sinx}\dfrac{\text{x}}{\text{2}}\text{cos}\dfrac{\text{x}}{\text{2}}}\\\end{align}

Therefore,

$\dfrac{\sqrt{\text{1+sinx}}\text{+}\sqrt{\text{1-sinx}}}{\sqrt{\text{1+sinx}}\text{-}\sqrt{\text{1-sinx}}}\text{=cot}\dfrac{\text{x}}{\text{2}}$ .                                                 …… (2)

So, from the equations (1) and (2) we obtain,

\begin{align}&\text{y=co}{{\text{t}}^{\text{-1}}}\left(\text{co}{{\text{t}}^{\dfrac{\text{x}}{\text{2}}}}\right)\\&\Rightarrow\text{y=}\dfrac{\text{x}}{\text{2}}\\\end{align}

Now, differentiating both sides with respect to  $\text{x}$  gives

$\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{1}}{\text{2}}\dfrac{\text{d}}{\text{dx}}\text{(x)}$

Hence,  $\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{1}}{\text{2}}$ .

7. $\text{y=(logx}{{\text{)}}^{\text{logx}}}\text{,}\,\,\,\text{x}>1$ .

Ans: The given function is  $\text{y=(logx}{{\text{)}}^{\text{logx}}}$ .

First, take the logarithm on both sides of the function.

$\text{logy=logx }\!\!\times\!\!\text{ log(logx)}$ .

Now, differentiating both sides with respect to  $\text{x}$  gives

\begin{align}&\dfrac{\text{1}}{\text{y}}\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{d}}{\text{dx}}\left[\text{logx}\!\!\times\!\!\text{log(logx)}\right]\\&\Rightarrow \dfrac{\text{1}}{\text{y}}\dfrac{\text{dy}}{\text{dx}}\text{=log(logx) }\!\!\times\!\!\text{}\dfrac{\text{d}}{\text{dx}}\text{(logx)+}\dfrac{\text{d}}{\text{dx}}\left[\text{log(logx)}\right]\\&\Rightarrow\dfrac{\text{dy}}{\text{dx}}\text{=y}\left[\text{log(logx)}\!\!\times\!\!\text{}\dfrac{\text{1}}{\text{x}}\text{+logx }\!\!\times\!\!\text{ }\dfrac{\text{1}}{\text{logx}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(logx)}\right]\\&\Rightarrow\dfrac{\text{dy}}{\text{dx}}\text{=y}\left[\dfrac{\text{1}}{\text{x}}\text{log(logx)+}\dfrac{\text{1}}{\text{x}} \right] \\ \end{align}

Hence,  $\dfrac{\text{dy}}{\text{dx}}\text{=(logx}{{\text{)}}^{\text{logx}}}\left[ \dfrac{\text{1}}{\text{x}}\text{+}\dfrac{\text{log(logx)}}{\text{x}} \right]$ .

8. $\mathbf{y=cos(acosx+bsinx)}$ , for some constants  $\text{a}$  and  $\mathbf{b}$ .

Ans: The given function is  $\text{y=cos(acosx+bsinx)}$.

Now, differentiating both sides with respect to  $\text{x}$  by using the chain rule of derivatives gives

\begin{align}&\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{d}}{\text{dx}}\text{cos(acosx+bsinx)}\\&\Rightarrow\dfrac{\text{dy}}{\text{dx}}\text{=-sin(acosx+bsinx)}\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(acosx+bsinx)} \\ & =\text{-sin(acosx+bsinx)}\!\!\times\!\!\text{}\left[\text{a(-sinx)+bcosx}\right] \\ \end{align}

Hence,  $\dfrac{\text{dy}}{\text{dx}}=\text{(asinx-bcosx) }\!\!\times\!\!\text{ sin(acosx+bsinx)}$ .

9. $\mathbf{y=(sinx-cosx}{{\mathbf{)}}^{\mathbf{(sinx-cosx)}}}\mathbf{,}\,\,\,\dfrac{\mathbf{\pi }}{\mathbf{4}}<\mathbf{x}<\dfrac{\mathbf{3\pi }}{\mathbf{4}}$ .

Ans: The given function is  $\text{y=(sinx-cosx}{{\text{)}}^{\text{(sinx-cosx)}}}$.

First, take the logarithm on both sides of the function.

\begin{align}& \text{logy=log}\left[ {{\text{(sinx-cosx)}}^{\text{(sinx-cosx)}}} \right] \\ &\Rightarrow\text{logy=(sinx-cosx)}\!\!\times\!\!\text{log(sinx-cosx)} \\ \end{align}

Now, differentiating both sides with respect to  $\text{x}$  gives

\begin{align}&\dfrac{\text{1}}{\text{y}}\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{d}}{\text{dx}}\left[ \text{(sinx-cosx) }\!\!\times\!\!\text{ log(sinx-cosx)} \right]\\&\Rightarrow\dfrac{\text{1}}{\text{y}}\dfrac{\text{dy}}{\text{dx}}\text{=log(sinx-cosx)}\!\!\times\!\!\text{}\dfrac{\text{d}}{\text{dx}}\text{(sinx-cosx)+(sinx-cosx) }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{log(sinx-cosx)} \\&\Rightarrow\dfrac{\text{1}}{\text{y}}\dfrac{\text{dy}}{\text{dx}}\text{=log(sinx-cosx)}\!\!\times\!\!\text{(cosx+sinx)+(sinx-cosx)}\!\!\times\!\!\text{}\dfrac{\text{1}}{\text{(sinx-cosx)}}\text{}\!\!\times\!\!\text{}\dfrac{\text{d}}{\text{dx}}\text{(sinx-cosx)}\\&\Rightarrow\dfrac{\text{dy}}{\text{dx}}\text{=(sinx-cosx}{{\text{)}}^{\text{(sinx-cosx)}}}\left[\left(\text{cosx+sinx}\right)\text{}\!\!\times\!\!\text{ log(sinx-cosx)+(cosx+sinx)} \right] \\ \end{align}

Hence, the required derivative is

$\dfrac{\text{dy}}{\text{dx}}\text{=(sinx-cosx}{{\text{)}}^{\text{(sinx-cosx)}}}\text{(cosx+sinx)}\left[ \text{1+log(sinx-cosx)} \right]$ .

10. $\mathbf{y=}{{\mathbf{x}}^{\mathbf{x}}}\mathbf{+}{{\mathbf{x}}^{\mathbf{a}}}\mathbf{+}{{\mathbf{a}}^{\mathbf{x}}}\mathbf{+}{{\mathbf{a}}^{\mathbf{a}}}$ , for some fixed  $\text{a}>0$  and  $\text{x}>0$ .

Ans: The given function is

$\text{y=}{{\text{x}}^{\text{x}}}\text{+}{{\text{x}}^{\text{a}}}\text{+}{{\text{a}}^{\text{x}}}\text{+}{{\text{a}}^{\text{a}}}$ .

Now, assume that  ${{\text{x}}^{\text{x}}}\text{=u}$ ,

${{\text{x}}^{\text{a}}}\text{=v}$ ,  ${{\text{a}}^{\text{x}}}\text{=w}$  and  ${{\text{a}}^{\text{a}}}\text{=s}$

Therefore, we have  $\text{y=u+v+w+s}$ .

So, differentiating both sides with respect to  $\text{x}$  gives

$\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{du}}{\text{dx}}\text{+}\dfrac{\text{dv}}{\text{dx}}\text{+}\dfrac{\text{dw}}{\text{dx}}\text{+}\dfrac{\text{ds}}{\text{dx}}$                                            …… (1)

Also,  $\text{u=}{{\text{x}}^{\text{x}}}$

\begin{align} & \Rightarrow \text{logu=log}{{\text{x}}^{\text{x}}} \\ & \Rightarrow \text{logu=xlogx} \\ \end{align}

Then, differentiating both sides with respect to  $\text{x}$  gives

\begin{align}&\dfrac{\text{1}}{\text{u}}\dfrac{\text{du}}{\text{dx}}\text{=logx}\text{.}\dfrac{\text{d}}{\text{dx}}\text{(x)+x}\text{.}\dfrac{\text{d}}{\text{dx}}\text{(logx)}\\&\Rightarrow\dfrac{\text{du}}{\text{dx}}\text{=u}\left[\text{logx}\text{.1+x}\text{.}\dfrac{\text{1}}{\text{x}} \right] \\ \end{align}

Thus,  $\dfrac{\text{du}}{\text{dx}}\text{=}{{\text{x}}^{\text{x}}}\text{ }\!\![\!\!\text{ logx+1 }\!\!]\!\!\text{ =}{{\text{x}}^{\text{x}}}\text{(1+logx)}$                           ……. (2)

Again,  $\text{v=}{{\text{x}}^{\text{a}}}$

Then, differentiating both sides with respect to  $\text{x}$  gives

$\dfrac{\text{du}}{\text{dx}}\text{=}\dfrac{\text{d}}{\text{dx}}\text{(}{{\text{x}}^{\text{a}}}\text{)}$

$\Rightarrow \dfrac{\text{dv}}{\text{dx}}\text{=a}{{\text{x}}^{\text{a-1}}}$                                                                  …… (3)

Also,  $\text{w=}{{\text{a}}^{\text{x}}}$

\begin{align}&\Rightarrow\text{logw=log}{{\text{a}}^{\text{x}}}\\&\Rightarrow \text{logw=xloga} \\ \end{align}

So, differentiating both sides with respect to  $\text{x}$  gives

\begin{align}&\dfrac{\text{1}}{\text{w}}\text{.}\dfrac{\text{dw}}{\text{dx}}\text{=loga}\text{.}\dfrac{\text{d}}{\text{dx}}\text{(x)}\\&\Rightarrow\dfrac{\text{dw}}{\text{dx}}\text{=wloga} \\ \end{align}

$\Rightarrow\dfrac{\text{dw}}{\text{dx}}\text{=}{{\text{a}}^{\text{x}}}\text{loga}$                                                          ……… (4)

and

$\text{s=}{{\text{a}}^{\text{a}}}$

Then differentiating both sides with respect to  $\text{x}$  gives

$\dfrac{\text{ds}}{\text{dx}}\text{=0}$ ,                                                                         ……(5)

as  $\text{a}$  is constant, and so  ${{\text{a}}^{\text{a}}}$  is also a constant.

Now, from the equations (1), (2), (3), (4), and (5) we have

$\dfrac{\text{dy}}{\text{dx}}\text{=}{{\text{x}}^{\text{x}}}\text{(1+logx)+a}{{\text{x}}^{\text{a-1}}}\text{+}{{\text{a}}^{\text{x}}}\text{loga+0}$

Hence,$\dfrac{\text{dy}}{\text{dx}}\text{=}{{\text{x}}^{\text{x}}}\text{(1+logx)+a}{{\text{x}}^{\text{a-1}}}\text{+}{{\text{a}}^{\text{x}}}\text{loga}$ .

11. $\mathbf{y=}{{\mathbf{x}}^{{{\mathbf{x}}^{\mathbf{2}}}\mathbf{-3}}}\mathbf{+(x-3}{{\mathbf{)}}^{{{\mathbf{x}}^{\mathbf{2}}}}}$ , for  $\text{x}>3$ .

Ans: The given function is

$\text{y=}{{\text{x}}^{{{\text{x}}^{\text{2}}}\text{-3}}}\text{+(x-3}{{\text{)}}^{{{\text{x}}^{\text{2}}}}}$ .

Now suppose that  $\text{u=}{{\text{x}}^{{{\text{x}}^{\text{2}}}\text{-3}}}$  and  $\text{v=(x-3}{{\text{)}}^{{{\text{x}}^{\text{2}}}}}$

Therefore,  $\text{y=u+v}$ .

Now, differentiating both sides with respect to  $\text{x}$  gives

$\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{du}}{\text{dx}}\text{+}\dfrac{\text{dv}}{\text{dx}}$                                                ……. (1)

Also,  $\text{u=}{{\text{x}}^{{{\text{x}}^{\text{2}}}\text{-3}}}$ .

Take the logarithm on both sides of the equation.

\begin{align}&\Rightarrow\text{logu=log(}{{\text{x}}^{{{\text{x}}^{\text{2}}}\text{-3}}}\text{)} \\&\Rightarrow\text{logu=(}{{\text{x}}^{\text{2}}}\text{-3)logx} \\ \end{align}

Differentiating both sides with respect to  $\text{x}$  gives

\begin{align}&\dfrac{\text{1}}{\text{u}}\dfrac{\text{du}}{\text{dx}}\text{=logx}\text{.}\dfrac{\text{d}}{\text{dx}}\text{(}{{\text{x}}^{\text{2}}}\text{-3)+(}{{\text{x}}^{\text{2}}}\text{-3)}\text{.}\dfrac{\text{d}}{\text{dx}}\text{(logx)}\\&\Rightarrow\dfrac{\text{1}}{\text{u}}\dfrac{\text{du}}{\text{dx}}\text{=logx}\text{.2x+(}{{\text{x}}^{\text{2}}}\text{-3)}\text{.}\dfrac{\text{1}}{\text{x}} \\ \end{align}

Hence,  $\dfrac{\text{du}}{\text{dx}}\text{=}{{\text{x}}^{{{\text{x}}^{\text{2}}}\text{-3}}}\text{.}\left[ \dfrac{{{\text{x}}^{\text{2}}}\text{-3}}{\text{x}}\text{+2 }\!\!\times\!\!\text{ logx} \right]$ .                            …… (2)

Again,  $\text{v=(x-3}{{\text{)}}^{{{\text{x}}^{\text{2}}}}}$ .

Take the logarithm on both sides of the equation.

\begin{align}&\Rightarrow\text{logv=log(x-3}{{\text{)}}^{{{\text{x}}^{\text{2}}}}}\\&\Rightarrow \text{logv=}{{\text{x}}^{\text{2}}}\text{log(x-3)} \\ \end{align}

Now, differentiating both sides with respect to  $\text{x}$  gives

\begin{align}&\dfrac{\text{1}}{\text{u}}\text{.}\dfrac{\text{dv}}{\text{dx}}\text{=log(x-3)}\text{.}\dfrac{\text{d}}{\text{dx}}\text{(}{{\text{x}}^{\text{2}}}\text{)+}{{\text{x}}^{\text{2}}}\text{.}\dfrac{\text{d}}{\text{dx}}\text{}\!\![\!\!\text{log(x-3)}\!\!]\!\!\text{}\\&\Rightarrow\dfrac{\text{1}}{\text{u}}\text{.}\dfrac{\text{dv}}{\text{dx}}\text{=log(x-3)}\text{.2x+}{{\text{x}}^{\text{2}}}\text{.}\dfrac{\text{1}}{\text{x-3}}\text{.}\dfrac{\text{d}}{\text{dx}}\text{(x-3)}\\&\Rightarrow\dfrac{\text{dv}}{\text{dx}}\text{=v}\text{.}\left[\text{2xlog(x-3)+}\dfrac{{{\text{x}}^{\text{2}}}}{\text{x-3}}\text{.1} \right] \\ \end{align}

Hence,$\dfrac{\text{dv}}{\text{dx}}\text{=(x-3}{{\text{)}}^{{{\text{x}}^{\text{2}}}}}\left[ \dfrac{{{\text{x}}^{\text{2}}}}{\text{x-3}}\text{+2xlog(x-3)} \right]$                                    …… (3)

Thus, from the equations (1), (2) and (3) we obtain

$\dfrac{\text{dy}}{\text{dx}}\text{=}{{\text{x}}^{{{\text{x}}^{\text{2}}}\text{-3}}}\left[ \dfrac{{{\text{x}}^{\text{2}}}\text{-3}}{\text{x}}\text{+2xlogx} \right]\text{+(x-3}{{\text{)}}^{{{\text{x}}^{\text{2}}}}}\left[ \dfrac{{{\text{x}}^{\text{2}}}}{\text{x-3}}\text{+2xlog(x-3)} \right]$ .

12. Find  $\dfrac{\mathbf{dy}}{\mathbf{dx}}$  if  $\mathbf{y=12(1-cost),x=10(t-sint),}\,\,-\dfrac{\mathbf{\pi }}{\mathbf{2}}<\mathbf{t}<\dfrac{\mathbf{\pi }}{\mathbf{2}}$ .

Ans: The given equations are  $\text{y=12(1-cost),}$                                 …… (1)

and  $\text{x=10(t-sint)}$                                                                    …… (2)

Then differentiating the equations (1) and (2) with respect to  $\text{x}$  gives

\begin{align}&\dfrac{\text{dx}}{\text{dt}}\text{=}\dfrac{\text{d}}{\text{dt}}\left[\text{10(t-sint)}\right]\text{=10}\!\!\times\!\!\text{}\dfrac{\text{d}}{\text{dt}}\text{(t-sint)=10(1-cost)}\\&\dfrac{\text{dy}}{\text{dt}}\text{=}\dfrac{\text{d}}{\text{dt}}\left[\text{12(1-cost)}\right]\text{=12}\!\!\times\!\!\text{}\dfrac{\text{d}}{\text{dt}}\text{(1-cost)=12 }\!\!\times\!\!\text{ }\left[ \text{0-(-sint)} \right]\text{=12sint} \\ \end{align}

Therefore, by dividing  $\dfrac{\text{dy}}{\text{dt}}$  by  $\dfrac{\text{dx}}{\text{dt}}$  we have,

$\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\left(\dfrac{\text{dy}}{\text{dt}}\right)}{\left(\dfrac{\text{dx}}{\text{dt}}\right)}\text{=}\dfrac{\text{12sint}}{\text{10(1-cost)}}\text{=}\dfrac{\text{12 }\!\!\times\!\!\text{ 2sin}\dfrac{\text{t}}{\text{2}}\text{ }\!\!\times\!\!\text{ cos}\dfrac{\text{t}}{\text{2}}}{\text{10}\!\!\times\!\!\text{2si}{{\text{n}}^{\text{2}}}\dfrac{\text{t}}{\text{2}}}$

Hence,  $\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{6}}{\text{5}}\text{cot}\dfrac{\text{t}}{\text{2}}$ .

13. Find  $\dfrac{\text{dy}}{\text{dx}}$ , if  $\text{y=si}{{\text{n}}^{\text{-1}}}\text{x+si}{{\text{n}}^{\text{-1}}}\sqrt{\text{1-}{{\text{x}}^{\text{2}}}}\text{,}\,\,\,\text{0}<\text{x}<\text{1}$ .

Ans: The given equation is  $\text{y=si}{{\text{n}}^{\text{-1}}}\text{x+si}{{\text{n}}^{\text{-1}}}\sqrt{\text{1-}{{\text{x}}^{\text{2}}}}$ .

Differentiating both sides of the equation with respect to  $\text{x}$  gives

\begin{align}&\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{d}}{\text{dx}}\left[\text{si}{{\text{n}}^{\text{-1}}}\text{x+si}{{\text{n}}^{\text{-1}}}\sqrt{\text{1-}{{\text{x}}^{\text{2}}}}\right]\\&\Rightarrow\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{d}}{\text{dx}}\text{(si}{{\text{n}}^{\text{-1}}}\text{x)+}\dfrac{\text{d}}{\text{dx}}\left(\text{si}{{\text{n}}^{\text{-1}}}\sqrt{\text{1-}{{\text{x}}^{\text{2}}}} \right) \\ \end{align}

\begin{align}&\Rightarrow\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{1}}{\sqrt{\text{1-}{{\text{x}}^{\text{2}}}}}\text{+}\dfrac{\text{1}}{\sqrt{\text{1-}{{\left( \sqrt{\text{1-}{{\text{x}}^{\text{2}}}} \right)}^{2}}}}\text{}\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\left( \sqrt{\text{1-}{{\text{x}}^{\text{2}}}} \right) \\ &\Rightarrow\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{1}}{\sqrt{\text{1-}{{\text{x}}^{\text{2}}}}}\text{-}\dfrac{1}{\sqrt{1-1+{{\text{x}}^{2}}}}\dfrac{\text{1}}{\text{2 }\!\!\times\!\!\text{ }\sqrt{\text{1-}{{\text{x}}^{\text{2}}}}}\times 2\text{x}\\&\Rightarrow\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{1}}{\sqrt{\text{1-}{{\text{x}}^{\text{2}}}}}\text{-}\dfrac{\text{1}}{\sqrt{\text{1-}{{\text{x}}^{\text{2}}}}} \\ \end{align}

Hence,  $\dfrac{\text{dy}}{\text{dx}}\text{=0}$ .

14. If  $\mathbf{x}\sqrt{\mathbf{1+y}}\mathbf{+y}\sqrt{\mathbf{1+x}}\mathbf{=0}$, for  $-1$<$x$<$1$ ,  prove that

$\dfrac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=-}\dfrac{\mathbf{1}}{{{\mathbf{(1+x)}}^{\mathbf{2}}}}$ .

Ans: The given equation is

\begin{align}& \text{x}\sqrt{\text{1+y}}\text{+y}\sqrt{\text{1+x}}\text{=0} \\ & \Rightarrow \text{x}\sqrt{\text{1+y}}\text{=y}\sqrt{\text{1+x}} \\ \end{align}

Now, squaring both sides of the equation gives

\begin{align}& {{\text{x}}^{\text{2}}}\text{(1+y)=}{{\text{y}}^{\text{2}}}\text{(1+x)} \\ & \Rightarrow {{\text{x}}^{\text{2}}}\text{+}{{\text{x}}^{\text{2}}}\text{y=}{{\text{y}}^{\text{2}}}\text{+x}{{\text{y}}^{\text{2}}} \\ & \Rightarrow {{\text{x}}^{\text{2}}}\text{-}{{\text{y}}^{\text{2}}}\text{=x}{{\text{y}}^{\text{2}}}\text{-}{{\text{x}}^{\text{2}}}\text{y} \\ & \Rightarrow {{\text{x}}^{\text{2}}}\text{-}{{\text{y}}^{\text{2}}}\text{=xy(y-x)} \\ & \Rightarrow \text{(x+y)(x-y)=xy(y-x)} \\ & \therefore \text{x+y=-xy} \\ & \Rightarrow \text{(1+x)y=-x} \\ & \Rightarrow \text{y=}\dfrac{\text{-x}}{\text{(1+x)}} \\ \end{align}

Now, differentiating both sides of the equation with respect to  $\text{x}$  gives

$\dfrac{\text{dy}}{\text{dx}}\text{=-}\dfrac{\text{(1+x)}\dfrac{\text{d}}{\text{dx}}\text{(x)-x}\dfrac{\text{d}}{\text{dx}}\text{(1+x)}}{{{\text{(1+x)}}^{\text{2}}}}\text{=-}\dfrac{\text{(1+x)-x}}{{{\text{(1+x)}}^{\text{2}}}}$

Hence,  $\dfrac{\text{dy}}{\text{dx}}\text{=-}\dfrac{\text{1}}{{{\text{(1+x)}}^{\text{2}}}}$.

15. If  ${{\mathbf{(x-a)}}^{\mathbf{2}}}\mathbf{+(y-b}{{\mathbf{)}}^{\mathbf{2}}}\mathbf{=}{{\mathbf{c}}^{\mathbf{2}}}$ , for some constant  $\text{c}>0$ , prove that  $\dfrac{{{\left[ \mathbf{1+}{{\left( \dfrac{\mathbf{dy}}{\mathbf{dx}} \right)}^{\mathbf{2}}} \right]}^{\dfrac{\mathbf{3}}{\mathbf{2}}}}}{\dfrac{{{\mathbf{d}}^{\mathbf{2}}}\mathbf{y}}{\mathbf{d}{{\mathbf{x}}^{\mathbf{2}}}}}$  is a constant independent of  $\text{a}$  and  $\text{b}$ .

Ans: The given equation is  ${{\text{(x-a)}}^{\text{2}}}\text{+(y-b}{{\text{)}}^{\text{2}}}\text{=}{{\text{c}}^{\text{2}}}$  .

Differentiating both sides of the equation with respect to  $\text{x}$  gives  \begin{align}& \dfrac{\text{d}}{\text{dx}}\text{= }\!\![\!\!\text{ (x-a}{{\text{)}}^{\text{2}}}\text{ }\!\!]\!\!\text{ +}\dfrac{\text{d}}{\text{dx}}\text{ }\!\![\!\!\text{ (y-b}{{\text{)}}^{\text{2}}}\text{ }\!\!]\!\!\text{ =}\dfrac{\text{d}}{\text{dx}}\text{(}{{\text{c}}^{\text{2}}}\text{)} \\ & \Rightarrow \text{2(x-a)}\text{.}\dfrac{\text{d}}{\text{dx}}\text{(x-a)+2(y-b)}\text{.}\dfrac{\text{d}}{\text{dx}}\text{(y-b)=0} \\ & \Rightarrow \text{2(x-a)}\text{.1+2(y-b)}\text{.}\dfrac{\text{dy}}{\text{dx}}\text{=0} \\ \end{align}

Hence,  $\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{-(x-a)}}{\text{y-b}}$                                                  ..…... (1)

Again, differentiating both sides of the equation with respect to  $\text{x}$  gives

\begin{align}& \dfrac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{=}\dfrac{\text{d}}{\text{dx}}\left[ \dfrac{\text{-(x-a)}}{\text{y-b}} \right] \\ & \text{=-}\dfrac{\left[ \text{(y-b)}\text{.}\dfrac{\text{d}}{\text{dx}}\text{(x-a)-(x-a)}\text{.}\dfrac{\text{d}}{\text{dx}}\text{(y-b)} \right]}{{{\text{(y-b)}}^{\text{2}}}} \\ & \text{=-}\left[ \dfrac{\text{(y-b)-(x-a)}\text{.}\dfrac{\text{dy}}{\text{dx}}}{{{\text{(y-b)}}^{\text{2}}}} \right] \\ & \text{=-}\left[ \dfrac{\text{(y-b)-(x-a)}\text{.}\left\{ \dfrac{\text{-(x-a)}}{\text{y-b}} \right\}}{{{\text{(y-b)}}^{\text{2}}}} \right] \\ & \text{=-}\left[ \dfrac{{{\text{(y-b)}}^{\text{2}}}\text{+(x+a}{{\text{)}}^{\text{2}}}}{{{\text{(y-b)}}^{\text{2}}}} \right] \\ \end{align}

Therefore,

${{\left[ \dfrac{\text{1+}{{\left( \dfrac{\text{dy}}{\text{dx}} \right)}^{\text{2}}}}{\dfrac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}} \right]}^{\dfrac{\text{3}}{\text{2}}}}\text{=}\dfrac{{{\left[ \left( \text{1+}\dfrac{{{\text{(x-a)}}^{\text{2}}}}{{{\text{(y-b)}}^{\text{2}}}} \right) \right]}^{\dfrac{\text{3}}{\text{2}}}}}{\text{-}\left[ \dfrac{{{\text{(y-a)}}^{\text{2}}}\text{+(x-a}{{\text{)}}^{\text{2}}}}{{{\text{(y-a)}}^{\text{3}}}} \right]}\text{=}\dfrac{{{\left[ \dfrac{{{\text{(y-b)}}^{\text{2}}}\text{+(x-a}{{\text{)}}^{\text{2}}}}{{{\text{(y-b)}}^{\text{2}}}} \right]}^{\dfrac{\text{3}}{\text{2}}}}}{\text{-}\left[ \dfrac{{{\text{(y-a)}}^{\text{2}}}\text{+(x-a}{{\text{)}}^{\text{2}}}}{{{\text{(y-a)}}^{\text{3}}}} \right]}=\dfrac{{{\left[ \dfrac{{{\text{c}}^{\text{2}}}}{{{\text{(y-b)}}^{\text{2}}}} \right]}^{\dfrac{\text{3}}{\text{2}}}}}{\text{-}\dfrac{{{\text{c}}^{\text{2}}}}{{{\text{(y-b)}}^{\text{3}}}}}$

$\Rightarrow{{\left[\dfrac{\text{1+}{{\left(\dfrac{\text{dy}}{\text{dx}}\right)}^{\text{2}}}}{\dfrac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}}\right]}^{\dfrac{\text{3}}{\text{2}}}}\text{=}\dfrac{\dfrac{{{\text{c}}^{\text{2}}}}{{{\text{(y-b)}}^{\text{3}}}}}{\dfrac{{{\text{c}}^{\text{2}}}}{{{\text{(y-b)}}^{\text{3}}}}}=\text{-c}$ , is a constant, and is independent of  $\text{a}$  and  $\text{b}$ .

16. If  $\text{cosy=xcos(a+y)}$ , with  $\text{cosa}\ne\text{}\!\!\pm\!\!\text{ 1}$ , prove that  $\dfrac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\dfrac{\mathbf{co}{{\mathbf{s}}^{\mathbf{2}}}\mathbf{(a+y)}}{\mathbf{sina}}$ .

Ans: The given equation is  $\text{cosy=xcos(a+y)}$ .

Then, differentiating both sides of the equation with respect to  $\text{x}$  gives

\begin{align}&\dfrac{\text{d}}{\text{dx}}\text{}\!\![\!\!\text{cosy}\!\!]\!\!\text{=}\dfrac{\text{d}}{\text{dx}}\text{}\!\![\!\!\text{xcos(a+y)}\!\!]\!\!\text{}\\&\Rightarrow\text{-siny}\dfrac{\text{dy}}{\text{dx}}\text{=cos(a+y)}\text{.}\dfrac{\text{d}}{\text{dx}}\text{(x)+x}\text{.}\dfrac{\text{d}}{\text{dx}}\text{}\!\![\!\!\text{cos(a+y)}\!\!]\!\!\text{}\\&\Rightarrow\text{-siny}\dfrac{\text{dy}}{\text{dx}}\text{=cos(a+y)+x}\text{.}\!\![\!\!\text{-sin(a+y)}\!\!]\!\!\text{}\dfrac{\text{dy}}{\text{dx}} \\ \end{align}

$\Rightarrow \text{ }\!\![\!\!\text{ xsin(a+y)-siny }\!\!]\!\!\text{ }\dfrac{\text{dy}}{\text{dx}}\text{=cos(a+y)}$                                  …….. (1)

Since  $\text{cosy=xcos(a+y)}\Rightarrow\text{x=}\dfrac{\text{cosy}}{\text{cos(a+y)}}$ , so from the equation (1) gives

\begin{align}&\left[\dfrac{\text{cosy}}{\text{cos(a+y)}}\text{.sin(a+y)-siny}\right]\dfrac{\text{dy}}{\text{dx}}\text{=cos(a+y)}\\&\Rightarrow\text{}\!\![\!\!\text{cosy}\text{.sin(a+y)-siny}\text{.cos(a+y)}\!\!]\!\!\text{}\text{.}\dfrac{\text{dy}}{\text{dx}}\text{=co}{{\text{s}}^{\text{2}}}\text{(a+y)}\\&\Rightarrow\text{sin(a+y-y)}\dfrac{\text{dy}}{\text{dx}}\text{=co}{{\text{s}}^{\text{2}}}\text{(a+y)} \\ \end{align}

Hence, it has been proved that  $\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{co}{{\text{s}}^{\text{2}}}\text{(a+y)}}{\text{sina}}$ .

17. If   $\mathbf{x=a(cost+tsint)}$  and  $\mathbf{y=a(sint-tcost)}$ , find  $\dfrac{{{\mathbf{d}}^{\mathbf{2}}}\mathbf{y}}{\mathbf{d}{{\mathbf{x}}^{\mathbf{2}}}}$ .

Ans: The given equations are

$\text{x=a(cost+tsint)}$                                            …… (1)

and  $\text{y=a(sint-tcost)}$                                        …… (2)

Then, differentiating both sides of the equation (1) with respect to  $\text{x}$  gives

\begin{align}&\dfrac{\text{dx}}{\text{dt}}\text{=a}\left[\text{-sint+sint}\text{.}\dfrac{\text{d}}{\text{dx}}\text{(t)+t}\text{.}\dfrac{\text{d}}{\text{dt}}\text{(sint)}\right]\\&\text{=a}\left[ \text{-sint+sint+cost} \right]\text{=atcost} \\ \end{align}

Again, differentiating both sides of the equation (2) with respect to  $\text{x}$  gives

\begin{align}&\dfrac{\text{dy}}{\text{dt}}\text{=a}\text{.}\dfrac{\text{d}}{\text{dt}}\text{(sint-tcost)}\\&\text{a}\left[\text{cost-}\left\{\text{cost}\text{.}\dfrac{\text{d}}{\text{dt}}\text{(t)+t}\text{.}\dfrac{\text{d}}{\text{dt}}\text{(cost)}\right\}\right]\\&\text{a}\!\![\!\!\text{ cost- }\!\!\{\!\!\text{ cost-tsint }\!\!\}\!\!\text{ }\!\!]\!\!\text{=atsint}\\\end{align}

Therefore,

$\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\left(\dfrac{\text{dy}}{\text{dt}}\right)}{\left( \dfrac{\text{dx}}{\text{dx}} \right)}\text{=}\dfrac{\text{atsint}}{\text{atcost}}\text{=tant}$

Now, differentiating both sides with respect to  $\text{x}$  gives

$\dfrac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{=}\dfrac{\text{d}}{\text{dx}}\left(\dfrac{\text{dy}}{\text{dx}}\right)\text{=}\dfrac{\text{d}}{\text{dx}}\text{(tant)=se}{{\text{c}}^{\text{2}}}\text{t}\text{.}\dfrac{\text{dt}}{\text{dx}}\text{=se}{{\text{c}}^{\text{2}}}\text{t}\text{.}\dfrac{\text{1}}{\text{atcost}}$

Hence,  $\dfrac{{{\mathbf{d}}^{\mathbf{2}}}\mathbf{y}}{\mathbf{d}{{\mathbf{x}}^{\mathbf{2}}}}\text{=}\dfrac{\text{se}{{\text{c}}^{\text{3}}}\text{(t)}}{\text{at}}$ .

18. If  $\mathbf{f(x)=}{{\left| \mathbf{x} \right|}^{\mathbf{3}}}$ , show that  $\mathbf{{{f}'}'}\left( \mathbf{x} \right)$  exists for all real  $\text{x}$   and find it.

Ans:  Remember that, \left|\text{x}\right|\text{=}\left\{\begin{align}&\text{x,if}\,\text{x}\ge\text{0}\\&\text{-x,if}\,\text{x0} \\ \end{align} \right\}

Therefore, if  $\text{x}\ge \text{0,}$  then  $\text{ f(x)=}{{\left| \text{x} \right|}^{\text{3}}}\text{=}{{\text{x}}^{\text{3}}}$ .

Then,  $\text{{f}'}\,\text{(x)=3}{{\text{x}}^{\text{2}}}$  .

Differentiating both sides with respect to  $\text{x}$  gives

$\text{{{f}'}'}\,\text{(x)=6x}$ .

Now, if $\text{x0,}$then$\text{f(x)=}{{\left|\text{x}\right|}^{\text{3}}}\text{=(-}{{\text{x}}^{\text{3}}}\text{)=}{{\text{x}}^{\text{3}}}$ .

So,  $\text{{f}'}\,\text{(x)=3}{{\text{x}}^{\text{2}}}$ .

Therefore, differentiating both sides with respect to  $\text{x}$  gives

$\text{{{f}'}'}\,\text{(x)=6x}$ .

Hence, for  $\text{f(x)=}{{\left| \text{x} \right|}^{\text{3}}}\text{,}$   $\text{{{f}'}'}\left( \text{x} \right)$  exists for all real values of  $\text{x}$  and is provided as

\text{{{f}'}'(x)=}\left\{\begin{align}&\text{6x,if}\,\text{x}\ge\text{0}\\&\text{-6x,if}\,\text{x0} \\ \end{align} \right\}

19. Using the fact that  $\mathbf{sin(A+B)=sinAcosB+cosAsinB}$  and the differentiation, obtain the sum formula for cosines.

Ans:  The given sum formula is  $\text{sin(A+B)=sinAcosB+cosAsinB}$ .

Now, differentiating both sides with respect to  $\text{x}$  gives

$\dfrac{\text{d}}{\text{dx}}\left[\text{sin(A+B)}\right]\text{=}\dfrac{\text{d}}{\text{dx}}\text{(sinAcosB)+}\dfrac{\text{d}}{\text{dx}}\text{(cosAsinB)}$

\begin{align}&\Rightarrow\text{cos(A+B)}\!\!\times\!\!\text{}\dfrac{\text{d}}{\text{dx}}\text{(A+B)=cosB}\!\!\times\!\!\text{}\dfrac{\text{d}}{\text{dx}}\text{(sinA)+sinA}\!\!\times\!\!\text{}\dfrac{\text{d}}{\text{dx}}\text{(cosB)+sinB}\!\!\times\!\!\text{}\dfrac{\text{d}}{\text{dx}}\text{(cosA)}\\&\text{+cosA}\!\!\times\!\!\text{}\dfrac{\text{d}}{\text{dx}}\text{(sinB)}\\&\Rightarrow\text{cos(A+B)}\!\!\times\!\!\text{}\dfrac{\text{d}}{\text{dx}}\text{(A+B)=cosB}\!\!\times\!\!\text{cosA}\dfrac{\text{d}}{\text{dx}}\text{+sinA(-sinB)}\dfrac{\text{dB}}{\text{dx}}\text{+sinB(-sinA)}\!\!\times\!\!\text{}\dfrac{\text{dA}}{\text{dx}}\\&\text{+cosAcosB}\dfrac{dB}{dx}\\&\Rightarrow\text{cos(A+B)}\left[\dfrac{\text{dA}}{\text{dx}}\text{+}\dfrac{\text{dB}}{\text{dx}}\right]\text{=(cosAcosB-sinAsinB)}\!\!\times\!\!\text{}\left[\dfrac{\text{dA}}{\text{dx}}\text{+}\dfrac{\text{dB}}{\text{dx}}\right]\\\end{align}

Hence the required sum formula for cosines is  $\text{cos(A+B)=cosAcosB-sinAsinB}$ .

20. Does there exist a function which is continuous everywhere but not differentiable at exactly two points?

Ans: Let take the function $\text{f}\left(\text{x}\right)=|\text{x}|+|\text{x}-1|$

Observe that, the function  $\text{f}$  is continuous everywhere, but not differentiable at $\text{x}=0$  and  $\text{x}=1$ .

21.If $\mathbf{y=}\left|\begin{matrix}\mathbf{f}\left(\mathbf{x}\right)&\mathbf{g}\left(\mathbf{x} \right) & \mathbf{h}\left( \mathbf{x} \right) \\\mathbf{l} & \mathbf{m} & \mathbf{n} \\\mathbf{a} & \mathbf{b} & \mathbf{c} \\\end{matrix} \right|$ , prove that$\dfrac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\left|\begin{matrix}\mathbf{{f}'}\left(\mathbf{x} \right)&\mathbf{{g}'}\left(\mathbf{x}\right)&\mathbf{{h}'}\left(\mathbf{x} \right) \\ \mathbf{l} & \mathbf{m} & \mathbf{n} \\ \mathbf{a} & \mathbf{b} & \mathbf{c} \\\end{matrix} \right|$  .

Ans:  The given function is  \text{y=}\left| \begin{align} & \text{f(x)}\,\,\,\text{g(x)}\,\,\,\text{h(x)} \\ & \text{ l}\ \ \ \ \ \ \text{m n} \\ & \text{ a b c } \\ \end{align} \right|

Evaluate the determinant.

$\text{y=(mc-nb)f(x)-(lc-na)g(x)+(lb-ma)h(x)}$ .

Now, differentiating both sides with respect to  $\text{x}$  gives

\begin{align}& \dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{d}}{\text{dx}}\text{ }\!\![\!\!\text{ (mc-nb)f(x) }\!\!]\!\!\text{ -}\dfrac{\text{d}}{\text{dx}}\text{ }\!\![\!\!\text{ (lc-na)g(x) }\!\!]\!\!\text{ +}\dfrac{\text{d}}{\text{dx}}\text{ }\!\![\!\!\text{ (lb-ma)h(x) }\!\!]\!\!\text{ } \\ & \text{=(mc-nb)f(x)-(lc-na)g(x)+(lb-ma)h(x)} \\ & =\left| \begin{matrix}\text{f}'\left(\text{x}\right)&\text{g}'\left(\text{x}\right)&\text{h}'\left(\text{x} \right) \\\text{l} & \text{m} & \text{n} \\\text{a} & \text{b} & \text{c} \\\end{matrix} \right| \\ \end{align}

Hence,  $\dfrac{\text{dy}}{\text{dx}}\text{=}\left|\begin{matrix}\text{f}'\left(\text{x}\right)&\text{g }'\left( \text{x} \right) & \text{h }'\left( \text{x} \right) \\\text{l} & \text{m} & \text{n} \\\text{a} & \text{b} & \text{c} \\\end{matrix} \right|$

22.If $\mathbf{y=}{{\mathbf{e}}^{\mathbf{aco}{{\mathbf{s}}^{\mathbf{-1}}}\mathbf{x}}}\text{, -1}\le \text{x}\le 1$ , show that $\mathbf{(1-}{{\mathbf{x}}^{\mathbf{2}}}\mathbf{)}\dfrac{{{\mathbf{d}}^{\mathbf{2}}}\mathbf{y}}{\mathbf{d}{{\mathbf{x}}^{\mathbf{2}}}}\mathbf{-x}\dfrac{\mathbf{dy}}{\mathbf{dx}}\mathbf{-}{{\mathbf{a}}^{\mathbf{2}}}\mathbf{y=0}$ .

Ans: The given equation is

$\text{y=}{{\text{e}}^{\text{aco}{{\text{s}}^{\text{-1}}}\text{x}}}$ .

Then take the logarithm on both sides of the equation.

\begin{align}&\text{logy=aco}{{\text{s}}^{\text{-1}}}\text{xloge}\\&\Rightarrow\text{logy=aco}{{\text{s}}^{\text{-1}}}\text{x} \\ \end{align}

Now, differentiating both sides with respect to  $\text{x}$  gives

\begin{align}&\dfrac{\text{1}}{\text{y}}\dfrac{\text{dy}}{\text{dx}}\text{=ax}\dfrac{\text{1}}{\sqrt{\text{1-}{{\text{x}}^{\text{2}}}}}\\&\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{-ax}}{\sqrt{\text{1-}{{\text{x}}^{\text{2}}}}} \\ \end{align}

Therefore, squaring both sides of the equation gives

\begin{align}&{{\left(\dfrac{\text{dy}}{\text{dx}}\right)}^{\text{2}}}\text{=}\dfrac{{{\text{a}}^{\text{2}}}{{\text{y}}^{\text{2}}}}{\text{1-}{{\text{x}}^{\text{2}}}}\\&\Rightarrow(\text{1-}{{\text{x}}^{\text{2}}}\text{)}{{\left(\dfrac{\text{dy}}{\text{dx}}\right)}^{\text{2}}}\text{=}{{\text{a}}^{\text{2}}}{{\text{y}}^{\text{2}}}\\&\Rightarrow\text{(1-}{{\text{x}}^{\text{2}}}\text{)}{{\left(\dfrac{\text{dy}}{\text{dx}}\right)}^{\text{2}}}\text{=}{{\text{a}}^{\text{2}}}{{\text{y}}^{\text{2}}} \\ \end{align}

Again, differentiating both sides with respect to  $\text{x}$  gives  \begin{align}&{{\left(\dfrac{\text{dy}}{\text{dx}}\right)}^{\text{2}}}\dfrac{\text{d}}{\text{dx}}\text{(1-}{{\text{x}}^{\text{2}}}\text{)+(1-}{{\text{x}}^{\text{2}}}\text{)}\!\!\times\!\!\text{}\dfrac{\text{d}}{\text{dx}}\left[{{\left(\dfrac{\text{dy}}{\text{dx}}\right)}^{\text{2}}}\right]\text{=}{{\text{a}}^{\text{2}}}\dfrac{\text{d}}{\text{dx}}\text{(}{{\text{y}}^{\text{2}}}\text{)}\\&\Rightarrow{{\left(\dfrac{\text{dy}}{\text{dx}}\right)}^{\text{2}}}\text{(-2x)+(1-}{{\text{x}}^{\text{2}}}\text{)}\!\!\times\!\!\text{2}\dfrac{\text{dy}}{\text{dx}}\text{}\!\!\times\!\!\text{}\dfrac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{=}{{\text{a}}^{\text{2}}}\text{}\!\!\times\!\!\text{2y}\!\!\times\!\!\text{}\dfrac{\text{dy}}{\text{dx}}\\&\Rightarrow\text{x}\dfrac{\text{dy}}{\text{dx}}\text{+(1-}{{\text{x}}^{\text{2}}}\text{)}\dfrac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{=}{{\text{a}}^{\text{2}}}\text{}\!\!\times\!\!\text{ y} \\ \end{align}

Hence, it is proved that  $\text{(1-}{{\text{x}}^{\text{2}}}\text{)}\dfrac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{-x}\dfrac{\text{dy}}{\text{dx}}\text{-}{{\text{a}}^{\text{2}}}\text{y=0}$

## Conclusion

Class 12 Maths Chapter 5 Miscellaneous Exercise Solutions PDF is important for understanding various concepts thoroughly. Miscellaneous Chapter 5 Class 12 covers diverse problems that require the application of multiple formulas and techniques. It's crucial to concentrate on comprehending the fundamental ideas behind every question as opposed to merely learning the answers by heart. To complete this task, keep in mind that you must comprehend the theory underlying each idea, practise frequently, and consult solved examples.

## Class 12 Maths Chapter 5: Exercises Breakdown

 Exercise Number of Questions Exercise 5.1 34 Questions and Solutions Exercise 5.2 10 Questions and Solutions Exercise 5.3 15 Questions and Solutions Exercise 5.4 10 Questions and Solutions Exercise 5.5 18 Questions and Solutions Exercise 5.6 11 Questions and Solutions Exercise 5.7 17 Questions and Solutions

## Chapter-Specific NCERT Solutions for Class 12 Maths

Given below are the chapter-wise NCERT Solutions for Class 12 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

## FAQs on NCERT Solutions for Class 12 Maths Miscellaneous Exercise Chapter 5- Continuity and Differentiability

1. How do I solve problems in Chapter 5 Continuity and Differentiability class 12 miscellaneous exercise?

The solutions would likely connect the concepts of continuity and differentiability. They might explain that a function can only be differentiable at a point if it's also continuous at that point. The solutions would then guide you through applying the rules of differentiation and checking for continuity conditions.

2. Are there solutions for problems with inverse trigonometric functions or exponential/logarithmic functions?

Yes, the miscellaneous exercise might include problems involving these newly introduced function types in Chapter 5. The NCERT solutions would guide how to differentiate these functions using the specific rules learned in the chapter. They might also help you check for the continuity of these functions using the established methods.

3. How can I improve my problem-solving skills for class 12 chapter 5 miscellaneous exercise questions?

Here are some tips:

• Solidify the foundation: Ensure you have a clear understanding of the definitions of continuity and differentiability, along with the rules for differentiating various function types covered in the chapter.

• Practice with different types of functions: Solve problems involving polynomial, trigonometric, exponential, logarithmic, and inverse trigonometric functions to gain experience applying the concepts.

• Focus on the thought process: When referring to the NCERT solutions, pay attention to the logical steps involved in solving a problem, not just memorizing the final answer.

4. Where can I find additional resources for practising class 12 chapter 5 miscellaneous exercise problems?

• NCERT Textbook: The textbook itself might provide solutions to some problems within the miscellaneous exercise section.

• Online Resources: Vedantu offers comprehensive solutions and explanations for these problems. You can find them through a web search using terms like "NCERT Solutions Class 12 Maths Chapter 5 Miscellaneous Exercise Continuity and Differentiability."

5. What topics are covered in the Continuity and Differentiability Class 12 chapter 5 Miscellaneous Exercise?

The Miscellaneous Exercise covers a mix of all the concepts learned in Chapter 5, including matrices, determinants, and their applications.

6. How many questions exist in the Miscellaneous Exercise Class 12 Chapter 5?

There are a total of 22 questions in the Miscellaneous Exercise of Chapter 5.

7. Are the questions in the class 12 chapter 5 Miscellaneous Exercise similar to the previous year's exam questions?

Yes, the Miscellaneous Exercise includes questions that are often similar to those asked in previous years' board exams.

8. Are there any tricky questions in the Miscellaneous Exercise Class 12 Chapter 5?

Yes, the Miscellaneous Exercise includes a few challenging questions that require a deeper understanding of the concepts.

9. Can I find the answers to the class 12 chapter 5 Miscellaneous Exercise online?

Yes, NCERT Solutions for the class 12 chapter 5 Miscellaneous Exercise are available online and can be downloaded from Vedantu’s website for free.