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

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

CBSE Class 12 Maths Chapter 5 Other Study Materials

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.