NCERT Solutions for Class 12 Maths Chapter 5 Continuity And Differentiability Ex 5.5

Exercise 5.5

1. Find the derivative of the function $\text{y=cosx }\!\!\times\!\!\text{ cos2x }\!\!\times\!\!\text{ cos3x}$ with respect to $\mathbf{x}$.

Ans.

The given function is $\text{y=cosx }\!\!\times\!\!\text{ cos2x }\!\!\times\!\!\text{ cos3x}$.

First, taking logarithm both sides of the equation give,

$\text{logy=log(cosx }\!\!\times\!\!\text{ cos2x }\!\!\times\!\!\text{ cos3x)}$

$\Rightarrow \text{logy=log(cosx)+log(cos2x)+log(cos3x)}$, by the property of logarithm.

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

$\dfrac{\text{1}}{\text{y}}\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{1}}{\text{cosx}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(cosx)+}\dfrac{\text{1}}{\text{cos2x}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(cos2x)+}\dfrac{\text{1}}{\text{cos3x}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(cos3x)} $ 

$\Rightarrow \dfrac{\text{dy}}{\text{dx}}\text{=y}\left[ \text{-}\dfrac{\text{sinx}}{\text{cosx}}\text{-}\dfrac{\text{sin2x}}{\text{cos2x}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(2x)-}\dfrac{\text{sin3x}}{\text{cos3x}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(3x)} \right] $ 

Therefore,

$\dfrac{\text{dy}}{\text{dx}}\text{=-cos }\!\!\times\!\!\text{ cos2x }\!\!\times\!\!\text{ cos3x}\left[ \text{tanx+2tan2x+3tan3x} \right]$.

 2. Find the derivative of the function  $\mathbf{y=}\sqrt{\dfrac{\mathbf{(x-1)(x-2)}}{\mathbf{(x-3)(x-4)(x-5)}}}$  with respect to $\mathbf{x}$.

Ans.

The given function is $\text{y=}\sqrt{\dfrac{\text{(x-1)(x-2)}}{\text{(x-3)(x-4)(x-5)}}}$.

First taking logarithm both sides of the equation give

$\text{logy=log}\sqrt{\dfrac{\text{(x-1)(x-2)}}{\text{(x-3)(x-4)(x-5)}}} $ 

$\Rightarrow \text{logy=}\dfrac{\text{1}}{\text{2}}\text{log}\left[ \dfrac{\text{(x-1)(x-2)}}{\text{(x-3)(x-4)(x-5)}} \right] $ 

$\Rightarrow \text{logy=}\dfrac{\text{1}}{\text{2}}\left[ \text{log }\!\!\{\!\!\text{ (x-1)(x-2) }\!\!\}\!\!\text{ -log }\!\!\{\!\!\text{ (x-3)(x-4)(x-5) }\!\!\}\!\!\text{ } \right] $ 

$\Rightarrow \text{logy=}\dfrac{\text{1}}{\text{2}}\text{ }\!\![\!\!\text{ log(x-1)+log(x-2)-log(x-3)-log(x-4)-log(x-5) }\!\!]\!\!\text{ } $ 

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

$\dfrac{\text{dy}}{\text{dx}}=\dfrac{\text{1}}{\text{2}}\dfrac{\text{d}}{\text{dx}}\text{ }\!\![\!\!\text{ log(x-1)+log(x-2)-log(x-3)-log(x-4)-log(x-5) }\!\!]\!\!\text{ }$.

$\dfrac{\text{1}}{\text{y}}\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{1}}{\text{2}}\left[ \dfrac{\text{1}}{\text{x-1}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(x-1)+}\dfrac{\text{1}}{\text{x-2}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(x-2)-}\dfrac{\text{1}}{\text{x-3}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(x-3)-}\dfrac{\text{1}}{\text{x-4}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(x-4)} \right. $ 

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left. \text{-}\dfrac{\text{1}}{\text{x-5}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(x-5)} \right] $ 

$\Rightarrow \dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{y}}{\text{2}}\left( \dfrac{\text{1}}{\text{x-1}}\text{+}\dfrac{\text{1}}{\text{x-2}}\text{+}\dfrac{\text{1}}{\text{x-3}}\text{+}\dfrac{\text{1}}{\text{x-4}}\text{+}\dfrac{\text{1}}{\text{x-5}} \right)$

Therefore,

$\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{1}}{\text{2}}\sqrt{\dfrac{\text{(x-1)(x-2)}}{\text{(x-3)(x-4)(x-5)}}}\left[ \dfrac{\text{1}}{\text{x-1}}\text{+}\dfrac{\text{1}}{\text{x-2}}\text{+}\dfrac{\text{1}}{\text{x-3}}\text{+}\dfrac{\text{1}}{\text{x-4}}\text{+}\dfrac{\text{1}}{\text{x-5}} \right]$.

3. Find the derivative of the function $\text{y=(logx}{{\text{)}}^{\text{cosx}}}$ with respect to $\mathbf{x}$.

Ans.

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

First, taking logarithm both sides of the equation give 

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

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

$\dfrac{\text{1}}{\text{y}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{d}}{\text{dx}}\text{(cosx) }\!\!\times\!\!\text{ log(logx)+cosx }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\left[ \text{log(logx)} \right]$

$\Rightarrow \dfrac{\text{1}}{\text{y}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{dy}}{\text{dx}}\text{=-sinxlog(logx)+cosx }\!\!\times\!\!\text{ }\dfrac{\text{1}}{\text{logx}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(logx)}$, by applying the chain rule.

$\Rightarrow \dfrac{\text{dy}}{\text{dx}}\text{=y}\left[ \text{-sinxlog(logx)+}\dfrac{\text{cosx}}{\text{logx}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{1}}{\text{x}} \right]$

Therefore,

$\dfrac{\text{dy}}{\text{dx}}\text{=(logx}{{\text{)}}^{\text{cosx}}}\left[ \dfrac{\text{cosx}}{\text{xlogx}}\text{-sin }\!\!\times\!\!\text{ log(logx)} \right]$.

4. Determine the derivative of the function $\text{y=}{{\text{x}}^{\mathbf{x}}}\text{-}{{\text{2}}^{\text{sinx}}}$ with respect to $\mathbf{x}$.

Ans.

The given function is $\text{y=}{{\text{x}}^{\text{x}}}\text{-}{{\text{2}}^{\text{sinx}}}$.

Now, let ${{\text{x}}^{\text{x}}}\text{=u}$                                                                      …… (1)

and ${{\text{2}}^{\text{sinx}}}\text{=v}$.                                                                           …… (2)

Therefore, $\text{y=u-v}$.                                                                 …… (3)

Then differentiating the equation (3) with respect to $\text{x}$ gives

$\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{du}}{\text{dx}}-\dfrac{\text{dv}}{\text{dx}}$                                                                       …… (4)

Now, taking logarithm both sides of the equation (1) give

$\text{log}\left( \text{u} \right)=\log \left( {{\text{x}}^{\text{x}}} \right) $ 

$\Rightarrow \log \text{u=xlogx} $ 

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

$\dfrac{\text{1}}{\text{u}}\dfrac{\text{du}}{\text{dx}}\text{=}\left[ \dfrac{\text{d}}{\text{dx}}\text{(x) }\!\!\times\!\!\text{ logx+x }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(logx)} \right] $ 

$\Rightarrow \dfrac{\text{du}}{\text{dx}}\text{=u}\left[ \text{1 }\!\!\times\!\!\text{ logx+x }\!\!\times\!\!\text{ }\dfrac{\text{1}}{\text{x}} \right] $ 

$\Rightarrow \dfrac{\text{du}}{\text{dx}}\text{=}{{\text{x}}^{\text{x}}}\text{(logx+1)} $ 

$\Rightarrow \dfrac{\text{du}}{\text{dx}}\text{=}{{\text{x}}^{\text{x}}}\text{(1+logx)}$                                                      …… (5) 

Now, taking logarithm both sides of the equation (2) give

$\text{log}\left( {{\text{2}}^{\text{sinx}}} \right)\text{=logv}$

$\Rightarrow \text{logv=sinx }\!\!\times\!\!\text{ log2}$.

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

$\dfrac{\text{1}}{\text{v}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{dv}}{\text{dx}}\text{=log2 }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(sinx}) $ 

$\Rightarrow \dfrac{\text{dv}}{\text{dx}}\text{=vlog2cosx} $ 

$\Rightarrow \dfrac{\text{dv}}{\text{dx}}\text{=}{{\text{2}}^{\text{sinx}}}\text{cosxlog2}$                                                 …… (6)      

Therefore, from the equation (4), (5) and (6) give

$\dfrac{\text{dy}}{\text{dx}}\text{=}{{\text{x}}^{\text{x}}}\text{(1+logx)-}{{\text{2}}^{\text{sinx}}}\text{cosxlog}2$.

5. Find the derivative of the function $\mathbf{y=(x+3}{{\mathbf{)}}^{\mathbf{2}}}{{\mathbf{(x+4)}}^{\mathbf{3}}}{{\mathbf{(x+5)}}^{\mathbf{4}}}$ with respect to $\mathbf{x}$.

Ans. 

The given function is $\text{y=(x+3}{{\text{)}}^{\text{2}}}{{\text{(x+4)}}^{\text{3}}}{{\text{(x+5)}}^{\text{4}}}$.

First, taking logarithm both sides of the equation give

$\text{logy=log}\left[ {{\text{(x+3)}}^{\text{2}}}{{\text{(x+4)}}^{\text{3}}}{{\text{(x+5)}}^{\text{4}}} \right]$

$\Rightarrow \text{logy=2log(x+3)+3log(x+4)+4log(x+5)}$

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

$\dfrac{\text{1}}{\text{y}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{dy}}{\text{dy}}\text{=2 }\!\!\times\!\!\text{ }\dfrac{\text{1}}{\text{x-3}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dz}}\text{(x+3)+3 }\!\!\times\!\!\text{ }\dfrac{\text{1}}{\text{x+4}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(x+4)+4 }\!\!\times\!\!\text{ }\dfrac{\text{1}}{\text{x+5}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(x+5)} $ 

$\Rightarrow \dfrac{\text{dy}}{\text{dx}}\text{=y}\left[ \dfrac{\text{2}}{\text{x+3}}\text{+}\dfrac{\text{3}}{\text{x+4}}\text{+}\dfrac{\text{4}}{\text{x+5}} \right] $ 

$\Rightarrow \dfrac{\text{dy}}{\text{dx}}\text{=(x+3}{{\text{)}}^{\text{2}}}{{\text{(x+4)}}^{\text{3}}}{{\text{(x+5)}}^{\text{4}}}\text{ }\!\!\times\!\!\text{ }\left[ \dfrac{\text{2}}{\text{x+3}}\text{+}\dfrac{\text{3}}{\text{x+4}}\text{+}\dfrac{\text{4}}{\text{x+5}} \right] $ 

$\Rightarrow \dfrac{\text{dy}}{\text{dx}}\text{=(x+3}{{\text{)}}^{\text{2}}}{{\text{(x+4)}}^{\text{3}}}{{\text{(x+5)}}^{\text{4}}}\text{ }\!\!\times\!\!\text{ }\left[ \dfrac{\text{2(x+4)(x+5)+3(x+3)(x+5)+4(x+3)(x+4)}}{\text{(x+3)(x+4)(x+5)}} \right] $ 

$\Rightarrow \dfrac{\text{dy}}{\text{dx}}\text{=(x+3}{{\text{)}}^{\text{2}}}{{\text{(x+4)}}^{\text{2}}}{{\text{(x+5)}}^{\text{2}}}\text{-}\left[ \text{2(}{{\text{x}}^{\text{2}}}\text{+9x+20)+3(}{{\text{x}}^{\text{2}}}\text{+9x+15)+4(}{{\text{x}}^{\text{2}}}\text{+7x+12)} \right] $ 

Therefore,

$\dfrac{\text{dy}}{\text{dx}}\text{=(x+3)(x+4}{{\text{)}}^{\text{2}}}{{\text{(x+5)}}^{\text{3}}}\text{(9}{{\text{x}}^{\text{2}}}\text{+70x+133)}$.

6. Find the derivative of the function $\mathbf{y=}{{\left( \mathbf{x+}\dfrac{\mathbf{1}}{\mathbf{x}} \right)}^{\mathbf{x}}}\mathbf{+}{{\mathbf{x}}^{\left( \mathbf{1+}\dfrac{\mathbf{1}}{\mathbf{x}} \right)}}$ with respect to $\mathbf{x}$.

Ans.

The given function is $\text{y=}{{\left( \text{x+}\dfrac{\text{1}}{\text{x}} \right)}^{\text{x}}}\text{+}{{\text{x}}^{\left( \text{1+}\dfrac{\text{1}}{\text{x}} \right)}}$.

First, let $\text{u=}{{\left( \text{x+}\dfrac{\text{1}}{\text{x}} \right)}^{\text{x}}}$and $\text{v=}{{\text{x}}^{\left( \text{1+}\dfrac{\text{1}}{\text{x}} \right)}}$

Therefore, $\text{y=u+v}$.                                         …… (1)

Differentiating the equation (1) both sides with respect to $\text{x}$ give

$\Rightarrow \dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{du}}{\text{dx}}\text{=}\dfrac{\text{dv}}{\text{dx}}$ …... (2)

Now, $\text{u=}{{\left( \text{x+}\dfrac{\text{1}}{\text{x}} \right)}^{\text{x}}}$

$\Rightarrow \text{logu=log}{{\left( \text{x+}\dfrac{\text{1}}{\text{x}} \right)}^{\text{x}}}$

$\Rightarrow \text{logu=xlog}\left( \text{x+}\dfrac{\text{1}}{\text{x}} \right)$

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

$\dfrac{\text{1}}{\text{u}}\dfrac{\text{du}}{\text{dx}}\text{=}\dfrac{\text{d}}{\text{dx}}\text{(x) }\!\!\times\!\!\text{ log}\left( \text{x+}\dfrac{\text{1}}{\text{x}} \right)\text{+x }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\left[ \text{log}\left( \text{x+}\dfrac{\text{1}}{\text{x}} \right) \right] $ 

$\Rightarrow \dfrac{\text{1}}{\text{u}}\dfrac{\text{du}}{\text{dx}}\text{=1 }\!\!\times\!\!\text{ log}\left( \text{x+}\dfrac{\text{1}}{\text{x}} \right)\text{+x }\!\!\times\!\!\text{ }\dfrac{\text{1}}{\left( \text{x+}\dfrac{\text{1}}{\text{x}} \right)}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\left( \text{x+}\dfrac{\text{1}}{\text{x}} \right) $ 

$\Rightarrow \dfrac{\text{du}}{\text{dx}}\text{=u}\left[ \text{log}\left( \text{x+}\dfrac{\text{1}}{\text{x}} \right)\text{+}\dfrac{\text{x}}{\left( \text{x+}\dfrac{\text{1}}{\text{x}} \right)}\text{ }\!\!\times\!\!\text{ }\left( \text{x+}\dfrac{\text{1}}{{{\text{x}}^{\text{2}}}} \right) \right] $ 

$\Rightarrow \dfrac{\text{du}}{\text{dx}}\text{=}{{\left( \text{x+}\dfrac{\text{1}}{\text{x}} \right)}^{\text{x}}}\left[ \text{log}\left( \text{x+}\dfrac{\text{1}}{\text{x}} \right)\text{+}\dfrac{\left( \text{x-}\dfrac{\text{1}}{\text{x}} \right)}{\left( \text{x+}\dfrac{\text{1}}{\text{x}} \right)} \right] $ 

$\Rightarrow \dfrac{\text{du}}{\text{dx}}\text{=}{{\left( \text{x+}\dfrac{\text{1}}{\text{x}} \right)}^{\text{x}}}\left[ \text{log}\left( \text{x+}\dfrac{\text{1}}{\text{x}} \right)\text{+}\dfrac{{{\text{x}}^{\text{2}}}\text{+1}}{{{\text{x}}^{\text{2}}}\text{-1}} \right] $

$\Rightarrow \dfrac{\text{du}}{\text{dx}}\text{=}{{\left( \text{x+}\dfrac{\text{1}}{\text{x}} \right)}^{\text{2}}}\left[ \dfrac{{{\text{x}}^{\text{2}}}\text{+1}}{{{\text{x}}^{\text{2}}}\text{-1}}\text{+log}\left( \text{x+}\dfrac{\text{1}}{\text{x}} \right) \right]$ …… (3)

Also, $\text{v=}{{\text{x}}^{\left( \text{x+}\dfrac{\text{1}}{\text{x}} \right)}}$

 $\Rightarrow \text{logv=log}\left[ {{\text{x}}^{{{\text{x}}^{\left( \text{x+}\dfrac{\text{1}}{\text{x}} \right)}}}} \right] $ 

$\Rightarrow \text{logv=}\left( \text{x+}\dfrac{\text{1}}{\text{x}} \right)\text{logx} $ 

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

$\dfrac{\text{1}}{\text{v}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{dv}}{\text{dx}}\text{=}\left[ \dfrac{\text{d}}{\text{dx}}\left( \text{1+}\dfrac{\text{1}}{\text{x}} \right) \right]\text{ }\!\!\times\!\!\text{ logx+}\left( \text{1+}\dfrac{\text{1}}{\text{x}} \right)\text{ }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{logx} $ 

$\Rightarrow \dfrac{\text{1}}{\text{v}}\dfrac{\text{dv}}{\text{dx}}\text{=-}\dfrac{\text{logx}}{{{\text{x}}^{\text{2}}}}\text{+}\dfrac{\text{1}}{\text{x}}\text{+}\dfrac{\text{1}}{{{\text{x}}^{\text{2}}}} $ 

$\Rightarrow \dfrac{\text{dv}}{\text{dx}}\text{=v}\left[ \dfrac{\text{-logx+x+1}}{{{\text{x}}^{\text{2}}}} \right]$           ……. (4)

Hence, from the equations (2), (3) and (4), give

$\dfrac{\text{dy}}{\text{dx}}\text{=}{{\left( \text{x+}\dfrac{\text{1}}{\text{x}} \right)}^{\text{x}}}\left[ \dfrac{{{\text{x}}^{\text{2}}}\text{-1}}{{{\text{x}}^{\text{2}}}\text{+1}}\text{+log}\left( \text{x+}\dfrac{\text{1}}{\text{x}} \right) \right]\text{+}{{\text{x}}^{\left( \text{x+}\dfrac{\text{1}}{\text{x}} \right)}}\left( \dfrac{\text{x+1-logx}}{{{\text{x}}^{\text{2}}}} \right)$.

7. Determine derivative of the function $\mathbf{y=(logx}{{\mathbf{)}}^{\mathbf{x}}}\mathbf{+}{{\mathbf{x}}^{\mathbf{logx}}}$ with respect to $\mathbf{x}$.

Ans.

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

Then, let $\text{u=(logx}{{\text{)}}^{\text{x}}}$ and $\text{v=}{{\text{x}}^{\text{logx}}}$.

Therefore, $\text{y=u+v}$.

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

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

Now, $\text{u=(logx}{{\text{)}}^{\text{x}}}$

$\Rightarrow \text{logu=log}\left[ {{\text{(logx)}}^{\text{x}}} \right] $ 

$\Rightarrow \text{logu=xlog(logx)} $ 

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

$\dfrac{\text{1}}{\text{u}}\dfrac{\text{du}}{\text{dx}}\text{=}\dfrac{\text{d}}{\text{dx}}\text{(x) }\!\!\times\!\!\text{ log(logx)+x }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\left[ \text{log(logx)} \right] $ 

$\Rightarrow \dfrac{\text{du}}{\text{dx}}\text{=u}\left[ \text{1 }\!\!\times\!\!\text{ log(logx)+x }\!\!\times\!\!\text{ }\dfrac{\text{1}}{\text{logx}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(logx)} \right] $ 

$\Rightarrow \dfrac{\text{du}}{\text{dx}}\text{=(logx}{{\text{)}}^{\text{x}}}\left[ \text{log(logx)+}\dfrac{\text{x}}{\text{logx}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{1}}{\text{x}} \right] $ 

$\Rightarrow \dfrac{\text{du}}{\text{dx}}\text{=(logx}{{\text{)}}^{\text{x}}}\left[ \text{log(logx)+}\dfrac{\text{1}}{\text{logx}} \right] $ 

$\Rightarrow \dfrac{\text{du}}{\text{dx}}\text{=(logx}{{\text{)}}^{\text{x}}}\text{=}\left[ \dfrac{\text{log(logx) }\!\!\times\!\!\text{ logx+1}}{\text{logx}} \right] $ 

$\dfrac{\text{du}}{\text{dx}}\text{=(logx}{{\text{)}}^{\text{x-1}}}\left[ \text{1+logx }\!\!\times\!\!\text{ log(logx)} \right]$ ……. (2)

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

$\Rightarrow \log \text{v=log}\left( {{\text{x}}^{\text{logx}}} \right) $ 

$\Rightarrow \log \text{v=logxlogx=}{{\left( \log \text{x} \right)}^{2}} $
Differentiating both sides of the equation with respect to $\text{x}$ gives

$\dfrac{\text{1}}{\text{v}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{dx}}{\text{dx}}\text{=}\dfrac{\text{d}}{\text{dx}}\left[ {{\text{(logx)}}^{\text{2}}} \right] $ 

$\Rightarrow \dfrac{\text{1}}{\text{v}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{dx}}{\text{dx}}\text{=2(logx) }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(logx)} $ 

$\Rightarrow \dfrac{\text{dv}}{\text{dx}}\text{=2}{{\text{x}}^{\text{logx}}}\dfrac{\text{logx}}{\text{x}} $ 

$\Rightarrow \dfrac{\text{dv}}{\text{dx}}\text{=2}{{\text{x}}^{\text{logx}}}\text{ }\!\!\times\!\!\text{ logx}$ ….…. (3)

Hence, from the equations (1), (2), and (3), gives

$\dfrac{\text{dy}}{\text{dx}}\text{=(logx}{{\text{)}}^{\text{x+1}}}\left[ \text{1+logx }\!\!\times\!\!\text{ log(logx)} \right]\text{+2}{{\text{x}}^{\text{logx-1}}}\text{ }\!\!\times\!\!\text{ logx}$.

8. Find the derivative of the function  $\mathbf{y=(sinx}{{\mathbf{)}}^{\mathbf{x}}}\mathbf{+si}{{\mathbf{n}}^{\mathbf{-1}}}\sqrt{\mathbf{x}}$ with respect to $\mathbf{x}$.

Ans.

The given function is $\text{y=(sinx}{{\text{)}}^{\text{x}}}\text{+si}{{\text{n}}^{\text{-1}}}\sqrt{\text{x}}$.

Now, let $\text{u=(sinx}{{\text{)}}^{\text{x}}}$ and $\text{v=si}{{\text{n}}^{\text{-1}}}\sqrt{\text{x}}$.

Therefore, $\text{y=u+v}$.

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

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

Now, $\text{u=(sinx}{{\text{)}}^{\text{x}}}$

$\Rightarrow \text{logu=xlog(sinx}{{\text{)}}^{\text{x}}} $ 

$\Rightarrow \text{logu=xlog(sinx)} $ 

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

$\dfrac{\text{1}}{\text{u}}\dfrac{\text{du}}{\text{dx}}\text{=}\dfrac{\text{d}}{\text{dx}}\text{(x) }\!\!\times\!\!\text{ log(sinx)+x }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\left[ \text{log(sinx)} \right] $ 

$\Rightarrow \dfrac{\text{du}}{\text{dx}}\text{=u}\left[ \text{1 }\!\!\times\!\!\text{ log(sinx)+x }\!\!\times\!\!\text{ }\dfrac{\text{1}}{\text{sinx}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(sinx)} \right] $ 

$\Rightarrow \dfrac{\text{du}}{\text{dx}}\text{=(sinx}{{\text{)}}^{\text{x}}}\left[ \text{log(sinx)+}\dfrac{\text{x}}{\text{sinx}}\text{ }\!\!\times\!\!\text{ cosx} \right] $ 

$\Rightarrow \dfrac{\text{du}}{\text{dx}}\text{=(sinx}{{\text{)}}^{\text{x}}}\text{(xcotx+logsinx)}$ ….… (2)

Again,$\text{v=si}{{\text{n}}^{\text{-1}}}\sqrt{\text{x}}$.

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

$\dfrac{\text{dv}}{\text{dx}}\text{=}\dfrac{\text{1}}{\sqrt{\text{1-(}\sqrt{\text{x}}{{\text{)}}^{\text{2}}}}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(}\sqrt{\text{x}}\text{)} $ 

$\Rightarrow \dfrac{\text{dv}}{\text{dx}}\text{=}\dfrac{\text{1}}{\sqrt{\text{1-x}}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{1}}{\text{2}\sqrt{\text{x}}} $ 

$\Rightarrow \dfrac{\text{dv}}{\text{dx}}\text{=}\dfrac{\text{1}}{\text{2}\sqrt{\text{x-}{{\text{x}}^{\text{2}}}}}$

Hence, from the equations (1), (2) and (3), gives

$\dfrac{\text{dv}}{\text{dx}}\text{=(sinx}{{\text{)}}^{\text{2}}}\text{(xcotx+logsinx)+}\dfrac{\text{1}}{\text{2}\sqrt{\text{x-}{{\text{x}}^{\text{2}}}}}$.

9. Find the derivative of the function $\mathbf{y=}{{\mathbf{x}}^{\mathbf{sinx}}}\mathbf{+(sinx}{{\mathbf{)}}^{\mathbf{cosx}}}$ with respect to $\mathbf{x}$.

Ans.

The given function is $\text{y=}{{\text{x}}^{\text{sinx}}}\text{+(sinx}{{\text{)}}^{\text{cosx}}}$.

Then, let $\text{u=}{{\text{x}}^{\text{sinx}}}$ and $\text{v=(sinx}{{\text{)}}^{\text{cosx}}}$.

Therefore, $\text{y=u+v}$.

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

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

Now, $\text{u=}{{\text{x}}^{\text{sinx}}}$

$\Rightarrow \text{logu=xlog(}{{\text{x}}^{\text{sinx}}}\text{)} $ 

$\Rightarrow \text{logu=sinxlogx} $ 

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

$\dfrac{\text{1}}{\text{u}}\dfrac{\text{du}}{\text{dx}}\text{=}\dfrac{\text{d}}{\text{dx}}\text{(sinx) }\!\!\times\!\!\text{ logx+sinx }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(logx)} $ 

$\Rightarrow \dfrac{\text{du}}{\text{dx}}\text{=u=}\left[ \text{cosxlogx+sinx }\!\!\times\!\!\text{ }\dfrac{\text{1}}{\text{x}} \right] $ 

$\Rightarrow \dfrac{\text{du}}{\text{dx}}\text{=}{{\text{x}}^{\text{sinx}}}\left[ \text{cosxlogx+}\dfrac{\text{sinx}}{\text{x}} \right]$                     ….... (2)

Again, $\text{v=(sinx}{{\text{)}}^{\text{cosx}}}$

$\Rightarrow \text{logv=log(sinx}{{\text{)}}^{\text{cosx}}} $ 

$\Rightarrow \text{logv=cosxlog(sinx)} $ 

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

$\dfrac{\text{1}}{\text{v}}\dfrac{\text{dv}}{\text{dx}}\text{=}\dfrac{\text{d}}{\text{dx}}\text{(cosx) }\!\!\times\!\!\text{ log(sinx)+cosx }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\left[ \text{log(sinx)} \right] $ 

$\Rightarrow \dfrac{\text{dv}}{\text{dx}}\text{=v}\left[ \text{-sinx }\!\!\times\!\!\text{ log(sinx)+cosx }\!\!\times\!\!\text{ }\dfrac{\text{1}}{\text{sinx}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(sinx)} \right] $ 

$\Rightarrow \dfrac{\text{du}}{\text{dx}}\text{=(sinx}{{\text{)}}^{\text{cosx}}}\left[ \text{-sinxlogsinx+cotxcosx} \right] $ 

$\Rightarrow \dfrac{\text{dv}}{\text{dx}}\text{=(sinx}{{\text{)}}^{\text{cosx}}}\text{ }\!\![\!\!\text{ cosxcotx+sinxlogsinx }\!\!]\!\!\text{ }$ …… (3)

Hence, from the equations (1), (2) and (3), gives

$\dfrac{\text{du}}{\text{dx}}\text{=}{{\text{x}}^{\text{sinx}}}\left( \text{cosxlogx+}\dfrac{\text{sinx}}{\text{x}} \right)\text{+(sinx}{{\text{)}}^{\text{cosx}}}\text{ }\!\![\!\!\text{ cosxcotx+sinxlogsinx }\!\!]\!\!\text{ }$.

10. Find the derivative function $\mathbf{y=}{{\mathbf{x}}^{\mathbf{xcosx}}}\mathbf{+}\dfrac{{{\mathbf{x}}^{\mathbf{2}}}\mathbf{+1}}{{{\mathbf{x}}^{\mathbf{2}}}\mathbf{-1}}$ with respect to $\mathbf{x}$.

Ans. 

The given function is $\text{y=}{{\text{x}}^{\text{xcosx}}}\text{+}\dfrac{{{\text{x}}^{\text{2}}}\text{+1}}{{{\text{x}}^{\text{2}}}\text{-1}}$.

First, let $\text{u=}{{\text{x}}^{\text{xcosx}}}$ and $\text{v=}\dfrac{{{\text{x}}^{\text{2}}}\text{+1}}{{{\text{x}}^{\text{2}}}\text{-1}}$.

Therefore, $\text{y=u+v}$.

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

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

Now, $\text{u=}{{\text{x}}^{\text{xcosx}}}$.

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

$\dfrac{\text{1}}{\text{u}}\dfrac{\text{du}}{\text{dx}}\text{=}\dfrac{\text{d}}{\text{dx}}\text{(x) }\!\!\times\!\!\text{ cosxlogx+x }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(cosx) }\!\!\times\!\!\text{ logx+xcosx }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(logx)} $ 

$\Rightarrow \dfrac{\text{du}}{\text{dx}}\text{=u}\left[ \text{1 }\!\!\times\!\!\text{ cosx }\!\!\times\!\!\text{ logx+x }\!\!\times\!\!\text{ (-sinx)logx+xcosx }\!\!\times\!\!\text{ }\dfrac{\text{1}}{\text{x}} \right] $ 

$\Rightarrow \dfrac{\text{du}}{\text{dx}}\text{=}{{\text{x}}^{\text{xcosx}}}\text{(cosxlogx-xsinxlogx+cosx})$ …… (2)

Again, $\text{v=}\dfrac{{{\text{x}}^{\text{2}}}\text{+1}}{{{\text{x}}^{\text{2}}}\text{-1}}$

$\Rightarrow \text{logv=log(}{{\text{x}}^{\text{2}}}\text{+1)-log(}{{\text{x}}^{\text{2}}}\text{-1)}$

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

$\dfrac{\text{1}}{\text{v}}\text{=}\dfrac{\text{dv}}{\text{dx}}\text{=}\dfrac{\text{2x}}{{{\text{x}}^{\text{2}}}\text{+1}}\text{-}\dfrac{\text{2x}}{{{\text{x}}^{\text{2}}}\text{-1}} $ 

$\Rightarrow \dfrac{\text{dv}}{\text{dx}}\text{=v}\left[ \dfrac{\text{2x(}{{\text{x}}^{\text{2}}}\text{-1)-2x(}{{\text{x}}^{\text{2}}}\text{+1)}}{\text{(}{{\text{x}}^{\text{2}}}\text{+1)(}{{\text{x}}^{\text{2}}}\text{-1)}} \right] $ 

$\Rightarrow \dfrac{\text{du}}{\text{dx}}\text{=}\dfrac{{{\text{x}}^{\text{2}}}\text{+1}}{{{\text{x}}^{\text{2}}}\text{-1}}\text{ }\!\!\times\!\!\text{ }\left[ \dfrac{\text{-4x}}{\text{(}{{\text{x}}^{\text{2}}}\text{+1)(}{{\text{x}}^{\text{2}}}\text{-1)}} \right] $ 

$\Rightarrow \dfrac{\text{dv}}{\text{dx}}\text{=}\dfrac{\text{-4x}}{{{\text{(}{{\text{x}}^{\text{2}}}\text{-1)}}^{\text{2}}}}$ …….. (3)

Hence, from the equations (1), (2) and (3), give

$\dfrac{\text{dv}}{\text{dx}}\text{=}{{\text{x}}^{\text{xcosx}}}\left[ \text{cosx(1+logx)-xsinxlogx} \right]\text{-}\dfrac{\text{4x}}{{{\text{(}{{\text{x}}^{\text{2}}}\text{-1)}}^{\text{2}}}}$.

11. Find the derivative of the function $\mathbf{y=(xcosx}{{\mathbf{)}}^{\mathbf{x}}}\mathbf{+(xsinx}{{\mathbf{)}}^{\dfrac{\mathbf{1}}{\mathbf{x}}}}$ with respect to $\mathbf{x}$.

Ans.

The given function is $\text{y=(xcosx}{{\text{)}}^{\text{x}}}\text{+(xsinx}{{\text{)}}^{\dfrac{\text{1}}{\text{x}}}}$.

Then, let $\text{u=(xcosx}{{\text{)}}^{\text{x}}}$ and $\text{v=(xsinx}{{\text{)}}^{\dfrac{\text{1}}{\text{x}}}}$.

Therefore, $\text{y=u+v}$.

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

Again, $\text{u=(cosx}{{\text{)}}^{\text{x}}}$

$\Rightarrow \text{logu=log(xcosx}{{\text{)}}^{\text{x}}} $ 

$\Rightarrow \text{logu=xlog(xcosx)} $ 

$\Rightarrow \text{logu=x }\!\![\!\!\text{ logx+logcosx }\!\!]\!\!\text{ } $ 

$\Rightarrow \text{logu=xlogx+xlogcosx} $ 

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

$\dfrac{\text{1}}{\text{u}}\dfrac{\text{du}}{\text{dx}}\text{=}\dfrac{\text{d}}{\text{dx}}\text{(xlogx+xlogcosx)} $ 

$\Rightarrow \dfrac{\text{du}}{\text{dx}}\text{=u}\left[ \left\{ \text{logx }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(x)+x }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(logx)} \right\}\text{+}\left\{ \text{logcosx }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(x)+x }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(logcosx)} \right\} \right] $ 

$\Rightarrow \dfrac{\text{du}}{\text{dx}}\text{=(xcosx}{{\text{)}}^{\text{x}}}\left[ \left\{ \text{logx }\!\!\times\!\!\text{ 1+x }\!\!\times\!\!\text{ }\dfrac{\text{1}}{\text{x}} \right\}\text{+}\left\{ \text{logcosx-1+x }\!\!\times\!\!\text{ }\dfrac{\text{1}}{\text{cosx}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(cosx)} \right\} \right] $ 

$\Rightarrow \dfrac{\text{du}}{\text{dx}}\text{=(xcosx}{{\text{)}}^{\text{x}}}\left[ \left\{ \text{logx+1} \right\}\text{+}\left\{ \text{logcosx-1+}\dfrac{\text{x}}{\text{cosx}}\text{ }\!\!\times\!\!\text{ (-sinx)} \right\} \right] $ 

$\Rightarrow \dfrac{\text{du}}{\text{dx}}\text{=(xcosx}{{\text{)}}^{\text{x}}}\left[ \text{(logx+1)+(logcosx-xtanx)} \right] $ 

$\Rightarrow \dfrac{\text{du}}{\text{dx}}\text{=(xcosx}{{\text{)}}^{\text{x}}}\left[ \text{1-xtanx+(logx+logcosx)} \right]$

Therefore,

$\dfrac{\text{du}}{\text{dx}}\text{=(xcosx}{{\text{)}}^{\text{x}}}\left[ \text{1-xtanx+(logx(xcosx)} \right]$   …….. (2)

Again, $\text{v=(xsinx}{{\text{)}}^{\dfrac{\text{1}}{\text{x}}}}$

$\Rightarrow \text{logv=log(xsinx}{{\text{)}}^{\dfrac{\text{1}}{\text{x}}}} $ 

$\Rightarrow \text{logv=}\dfrac{\text{1}}{\text{x}}\text{log(xsinx)} $ 

$\Rightarrow \text{logv=}\dfrac{\text{1}}{\text{x}}\text{(logx+logsinx)} $ 

$\Rightarrow \text{logv=}\dfrac{\text{1}}{\text{x}}\text{logx+}\dfrac{\text{1}}{\text{x}}\text{logsinx} $ 

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

$\dfrac{\text{1}}{\text{v}}\dfrac{\text{dv}}{\text{dx}}\text{=}\dfrac{\text{d}}{\text{dx}}\left( \dfrac{\text{1}}{\text{x}}\text{logx} \right)\text{+}\dfrac{\text{d}}{\text{dx}}\left[ \dfrac{\text{1}}{\text{x}}\text{log(sinx)} \right] $ 

$\Rightarrow \dfrac{\text{1}}{\text{v}}\dfrac{\text{dv}}{\text{dx}}\text{=}\left[ \dfrac{\text{1}}{\text{x}}\text{logx }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\left( \dfrac{\text{1}}{\text{x}} \right)\text{+}\dfrac{\text{1}}{\text{x}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(logx)} \right]\text{+}\left[ \text{log(sinx) }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\left( \dfrac{\text{1}}{\text{x}} \right)\text{+}\dfrac{\text{1}}{\text{x}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\left\{ \text{(logsinx)} \right\} \right] $ 

$\Rightarrow \dfrac{\text{1}}{\text{v}}\dfrac{\text{dv}}{\text{dx}}\text{=}\left[ \dfrac{\text{1}}{\text{x}}\text{logx }\!\!\times\!\!\text{ }\left( \text{-}\dfrac{\text{1}}{{{\text{x}}^{\text{2}}}} \right)\text{+}\dfrac{\text{1}}{\text{x}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{1}}{\text{x}} \right]\text{+}\left[ \text{log(sinx) }\!\!\times\!\!\text{ }\left( \text{-}\dfrac{\text{1}}{{{\text{x}}^{\text{2}}}} \right)\text{+}\dfrac{\text{1}}{\text{x}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{1}}{\text{sinx}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(sinx)} \right] $ 

$\Rightarrow \dfrac{\text{1}}{\text{v}}\dfrac{\text{dv}}{\text{dx}}\text{=}\dfrac{\text{1}}{{{\text{x}}^{\text{2}}}}\text{(1-logx)+}\left[ \dfrac{\text{1-logx}}{{{\text{x}}^{\text{2}}}}\text{+}\dfrac{\text{1}}{\text{xsinx}}\text{ }\!\!\times\!\!\text{ cosx} \right] $ 

$\Rightarrow \dfrac{\text{1}}{\text{v}}\dfrac{\text{dv}}{\text{dx}}\text{=}\dfrac{\text{1}}{{{\text{x}}^{\text{2}}}}{{\text{(xsinx)}}^{\dfrac{\text{1}}{\text{x}}}}\text{+}\left[ \dfrac{\text{1-logx}}{{{\text{x}}^{\text{2}}}}\text{+}\dfrac{\text{-log(sinx)+xcotx}}{{{\text{x}}^{\text{2}}}} \right] $ 

$\Rightarrow \dfrac{\text{dv}}{\text{dx}}\text{=(xsinx}{{\text{)}}^{\dfrac{\text{1}}{\text{x}}}}\left[ \dfrac{1-\log x-\log (\sin x)+x\cot x}{{{\text{x}}^{\text{2}}}} \right] $ 

Therefore,

$\dfrac{\text{dv}}{\text{dx}}\text{=(xsinx}{{\text{)}}^{\dfrac{\text{1}}{\text{x}}}}\left[ \dfrac{\text{1-log(xsinx)+xcotx}}{{{\text{x}}^{\text{2}}}} \right]$ ……. (3)

Hence, from the equations (1), (2) and (3), gives

$\Rightarrow \dfrac{\text{dy}}{\text{dx}}\text{=(xcosx}{{\text{)}}^{\text{2}}}\left[ \text{1-xtanx+log(xcosx)} \right]\text{+(xsinx}{{\text{)}}^{\dfrac{\text{1}}{\text{x}}}}\left[ \dfrac{\text{1-log(xsinx)+xcotx}}{{{\text{x}}^{\text{2}}}} \right]$.

12. Determine $\dfrac{\mathbf{dy}}{\mathbf{dx}}$ from the equation ${{\mathbf{x}}^{\mathbf{y}}}\mathbf{+}{{\mathbf{y}}^{\mathbf{x}}}\mathbf{=1}$.

Ans.

The given function is ${{\text{x}}^{\text{y}}}\text{+}{{\text{y}}^{\text{x}}}\text{=1}$.

Then, let ${{\text{x}}^{\text{y}}}\text{=u}$ and ${{\text{y}}^{\text{x}}}\text{=v}$.

Therefore, $\text{u+v=1}$.

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

$\dfrac{\text{du}}{\text{dx}}\text{+}\dfrac{\text{dv}}{\text{dy}}\text{=0}$

Now, $\text{u=}{{\text{x}}^{\text{y}}}$                             ……. (1)

$\Rightarrow \text{logu=log(}{{\text{x}}^{\text{y}}}\text{)} $ 

$\Rightarrow \text{logu=ylogx} $ 

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

$\dfrac{\text{1}}{\text{u}}\dfrac{\text{du}}{\text{dx}}\text{=logx}\dfrac{\text{dy}}{\text{dx}}\text{+y }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(logx)}$

$\Rightarrow \dfrac{\text{du}}{\text{dx}}\text{=u}\left[ \text{logx}\dfrac{\text{dy}}{\text{dx}}\text{+y }\!\!\times\!\!\text{ }\dfrac{\text{1}}{\text{x}} \right]$

Therefore, $\dfrac{\text{du}}{\text{dx}}\text{=}{{\text{x}}^{\text{y}}}\left[ \text{logx}\dfrac{\text{dy}}{\text{dx}}\text{+}\dfrac{\text{y}}{\text{x}} \right]$ ………… (2)

Also, $\text{v=}{{\text{y}}^{\text{x}}}$

Taking logarithm both sides of the equation give

$\Rightarrow \text{logv=log(}{{\text{y}}^{\text{3}}}\text{)} $ 

$\Rightarrow \text{logv=xlogy} $ 

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

$\dfrac{\text{1}}{\text{v}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{dv}}{\text{dx}}\text{=logy }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(x)+x }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(logy)}$

$\Rightarrow \dfrac{\text{dv}}{\text{dx}}\text{=v}\left( \text{logy }\!\!\times\!\!\text{ 1+x }\!\!\times\!\!\text{ }\dfrac{\text{1}}{\text{y}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{dy}}{\text{dx}} \right)$

Therefore, $\dfrac{\text{dv}}{\text{dx}}\text{=}{{\text{y}}^{\text{x}}}\left( \text{logy+}\dfrac{\text{x}}{\text{y}}\dfrac{\text{dy}}{\text{dx}} \right)$ ……... (3)

So, from the equation (1), (2) and (3), gives

${{\text{x}}^{\text{y}}}\left( \text{logx}\dfrac{\text{dy}}{\text{dx}}\text{+}\dfrac{\text{y}}{\text{x}} \right)\text{+}{{\text{y}}^{\text{x}}}\left( \text{logy+}\dfrac{\text{x}}{\text{y}}\dfrac{\text{dy}}{\text{dx}} \right)\text{=0} $ 

$\Rightarrow \left( {{\text{x}}^{\text{2}}}\text{+logx+x}{{\text{y}}^{\text{y-1}}} \right)\dfrac{\text{dy}}{\text{dx}}\text{=-}\left( \text{y}{{\text{x}}^{\text{y-1}}}\text{+}{{\text{y}}^{\text{x}}}\text{logy} \right) $ 

Hence, $\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{y}{{\text{x}}^{\text{y-1}}}\text{+}{{\text{y}}^{\text{x}}}\text{logy}}{{{\text{x}}^{\text{y}}}\text{logx+x}{{\text{y}}^{\text{x-1}}}}$.

13. Determine $\dfrac{\mathbf{dy}}{\mathbf{dx}}$ from the equation ${{\text{y}}^{\text{x}}}\text{=}{{\text{x}}^{\text{y}}}$.

Ans.

The given equation is ${{\text{y}}^{\text{x}}}\text{=}{{\text{x}}^{\text{y}}}$.

Then, taking logarithm both sides of the equation give

$\text{xlogy=ylogx}$.

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

$\text{logy }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(x)+x }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(logy)=logx }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(y)+y }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(logx)} $ 

$\Rightarrow \text{logy }\!\!\times\!\!\text{ 1+x }\!\!\times\!\!\text{ }\dfrac{\text{1}}{\text{y}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{dy}}{\text{dx}}\text{=logx }\!\!\times\!\!\text{ }\dfrac{\text{dy}}{\text{dx}}\text{+y }\!\!\times\!\!\text{ }\dfrac{\text{1}}{\text{x}} $ 

$\Rightarrow \text{logy+}\dfrac{\text{x}}{\text{y}}\dfrac{\text{dy}}{\text{dx}}\text{=logx}\dfrac{\text{dy}}{\text{dx}}\text{+}\dfrac{\text{y}}{\text{x}} $ 

$\Rightarrow \left( \dfrac{\text{x}}{\text{y}}\text{-logx} \right)\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{y}}{\text{x}}\text{-logy} $ 

$\Rightarrow \left( \dfrac{\text{x-ylogx}}{\text{y}} \right)\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{y-xlogy}}{\text{x}} $ 

$\Rightarrow \left( \dfrac{\text{x-ylogx}}{\text{y}} \right)\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{y-xlogy}}{\text{x}} $ 

Therefore, $\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{y}}{\text{x}}\left( \dfrac{\text{y-xlogy}}{\text{x-ylogx}} \right)$.

14. Determine $\dfrac{\mathbf{dy}}{\mathbf{dx}}$ from the equation ${{\text{(cosx)}}^{\text{y}}}\text{=(cosy}{{\text{)}}^{\text{x}}}$.

Ans.

The given equation is ${{\text{(cosx)}}^{\text{y}}}\text{=(cosy}{{\text{)}}^{\text{x}}}$.

Then, taking logarithm both sides of the equation give

$\text{ylogcosx=xlogcosy}$.

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

$\text{logcosx }\!\!\times\!\!\text{ }\dfrac{\text{dy}}{\text{dx}}\text{+y }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(logcosx)=logcosy }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(x)+x }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(logcosy}) $ 

$\Rightarrow \text{logcosx}\dfrac{\text{dy}}{\text{dx}}\text{+}\dfrac{\text{y}}{\text{cosx}}\text{ }\!\!\times\!\!\text{ (-sinx)=logcosy+}\dfrac{\text{x}}{\text{cosy}}\text{(-siny) }\!\!\times\!\!\text{ }\dfrac{\text{dy}}{\text{dx}} $ 

$\Rightarrow \text{logcosx}\dfrac{\text{dy}}{\text{dx}}\text{-ytanx=logcosy-xtany}\dfrac{\text{dy}}{\text{dx}} $ 

$\Rightarrow \text{(logcosx+xtany)}\dfrac{\text{dy}}{\text{dx}}\text{=ytanx+logcosy} $ 

Therefore, $\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{ytanx+logcosy}}{\text{xtany+logcosx}}$.

15. Determine $\dfrac{\mathbf{dy}}{\mathbf{dx}}$ from the equation $\text{xy=}{{\text{e}}^{\text{(x-y)}}}$.

Ans. 

The given equation is $\text{xy=}{{\text{e}}^{\text{(x-y)}}}$.

Then, taking logarithm both sides of the equation give

$\text{log(xy)=log(}{{\text{e}}^{\text{x-y}}}\text{)} $ 

$\Rightarrow \text{logx+logy=(x-y)loge} $ 

$\Rightarrow \text{logx+logy=(x-y) }\!\!\times\!\!\text{ 1} $ 

$\Rightarrow \text{logx+logy=x-y} $ 

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

$\dfrac{\text{d}}{\text{dx}}\text{(logx)+}\dfrac{\text{d}}{\text{dx}}\text{(logy)=}\dfrac{\text{d}}{\text{dx}}\text{(x)-}\dfrac{\text{dy}}{\text{dx}} $ 

$\Rightarrow \dfrac{\text{1}}{\text{x}}\text{+}\dfrac{\text{1}}{\text{y}}\dfrac{\text{dy}}{\text{dx}}\text{=1-}\dfrac{\text{1}}{\text{x}} $ 

$\Rightarrow \left( \text{1+}\dfrac{\text{1}}{\text{y}} \right)\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{x-1}}{\text{x}} $ 

Therefore, $\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{y(x-1)}}{\text{x(x+1)}}$.

16. Determine the derivative of the following function $\mathbf{f}$ and hence evaluate $\mathbf{f'(1)}$. 

$\text{f(x)=(1+x)(1+}{{\text{x}}^{\text{2}}}\text{)(1+}{{\text{x}}^{\text{4}}}\text{)(1+}{{\text{x}}^{\text{8}}}\text{)}$.

Ans.

The given function is $\text{f(x)=(1+x)(1+}{{\text{x}}^{\text{2}}}\text{)(1+}{{\text{x}}^{\text{4}}}\text{)(1+}{{\text{x}}^{\text{8}}}\text{)}$.

By taking logarithm both sides of the equation give

$\text{logf(x)=log(1+x)+log(1+}{{\text{x}}^{\text{2}}}\text{)+log(1+}{{\text{x}}^{\text{4}}}\text{)+log(1+}{{\text{x}}^{\text{8}}}\text{)}$

Now, differentiating both sides of the equation with respect to $\text{x}$ gives $\dfrac{\text{1}}{\text{f(x)}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{ }\!\![\!\!\text{ f(x) }\!\!]\!\!\text{ =}\dfrac{\text{d}}{\text{dx}}\text{log(1+x)+}\dfrac{\text{d}}{\text{dx}}\text{log(1+}{{\text{x}}^{\text{2}}}\text{)+}\dfrac{\text{d}}{\text{dx}}\text{log(1+}{{\text{x}}^{\text{4}}}\text{)+}\dfrac{\text{d}}{\text{dx}}\text{log(1+}{{\text{x}}^{\text{8}}}\text{)} $ 

$\Rightarrow \dfrac{\text{1}}{\text{f(x)}}\text{ }\!\!\times\!\!\text{ f }\!\!'\!\!\text{ (x)=}\dfrac{\text{1}}{\text{1+x}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{1}}{\text{dx}}\text{(1+x)+}\dfrac{\text{1}}{\text{1+}{{\text{x}}^{\text{2}}}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{log(1+}{{\text{x}}^{\text{2}}}\text{)+}\dfrac{\text{1}}{\text{1+}{{\text{x}}^{\text{4}}}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{log(1+}{{\text{x}}^{\text{4}}}\text{)} $ 

$\text{+}\dfrac{\text{1}}{\text{1+}{{\text{x}}^{\text{8}}}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{log(1+}{{\text{x}}^{\text{8}}}\text{)} $ 

$\Rightarrow \text{f }\!\!'\!\!\text{ (x)=f(x)}\left[ \dfrac{\text{1}}{\text{1+x}}\text{+}\dfrac{\text{1}}{\text{1+}{{\text{x}}^{\text{2}}}}\text{ }\!\!\times\!\!\text{ 2x+}\dfrac{\text{1}}{\text{1+}{{\text{x}}^{\text{4}}}}\text{ }\!\!\times\!\!\text{ 4}{{\text{x}}^{\text{3}}}\text{+}\dfrac{\text{1}}{\text{1+}{{\text{x}}^{\text{8}}}}\text{ }\!\!\times\!\!\text{ 8}{{\text{x}}^{\text{7}}} \right] $ 

Therefore,

$\text{f }\!\!'\!\!\text{ (x)=(1+x)(1+}{{\text{x}}^{\text{2}}}\text{)(1+}{{\text{x}}^{\text{4}}}\text{)(1+}{{\text{x}}^{\text{8}}}\text{)}\left[ \dfrac{\text{1}}{\text{1+x}}\text{+}\dfrac{\text{2x}}{\text{1+}{{\text{x}}^{\text{2}}}}\text{+}\dfrac{\text{4}{{\text{x}}^{\text{3}}}}{\text{1+}{{\text{x}}^{\text{4}}}}\text{+}\dfrac{\text{8}{{\text{x}}^{\text{7}}}}{\text{1+}{{\text{x}}^{\text{8}}}} \right]$

So, 

$\text{f }\!\!'\!\!\text{ (1)=(1+1)(1+}{{\text{1}}^{\text{2}}}\text{)(1+}{{\text{1}}^{\text{4}}}\text{)(1+}{{\text{1}}^{\text{8}}}\text{)}\left[ \dfrac{\text{1}}{\text{1+1}}\text{+}\dfrac{\text{2 }\!\!\times\!\!\text{ 1}}{\text{1+}{{\text{1}}^{\text{2}}}}\text{+}\dfrac{\text{4 }\!\!\times\!\!\text{ }{{\text{1}}^{\text{3}}}}{\text{1+}{{\text{1}}^{\text{4}}}}\text{+}\dfrac{\text{8 }\!\!\times\!\!\text{ }{{\text{1}}^{\text{7}}}}{\text{1+}{{\text{1}}^{\text{8}}}} \right] $ 

$\text{=2 }\!\!\times\!\!\text{ 2 }\!\!\times\!\!\text{ 2 }\!\!\times\!\!\text{ 2}\left[ \dfrac{\text{1}}{\text{2}}\text{+}\dfrac{\text{2}}{\text{2}}\text{+}\dfrac{\text{4}}{\text{2}}\text{+}\dfrac{\text{8}}{\text{2}} \right] $ 

$\text{=16 }\!\!\times\!\!\text{ }\left( \dfrac{\text{1+2+4+8}}{\text{2}} \right) $ 

$\text{=16 }\!\!\times\!\!\text{ }\dfrac{\text{15}}{\text{2}}\text{=120} $ 

Hence, $\text{{f}'}\left( \text{1} \right)\text{=120}$.

17. Differentiate the function $\mathbf{y=(}{{\mathbf{x}}^{\mathbf{2}}}\mathbf{-5x+8)(}{{\mathbf{x}}^{\mathbf{3}}}\mathbf{+7x+9)}$ in three ways as described below. Also, verify whether all the answers are the same. 

(a) By using product rules. 

Ans.

The given function is $\text{y=}\left( {{\text{x}}^{\text{2}}}\text{-5x+8} \right)\left( {{\text{x}}^{\text{3}}}\text{+7x+9} \right)$.

Now, let consider $\text{u=(}{{\text{x}}^{\text{2}}}\text{-5x+8)}$ and $\text{v=(}{{\text{x}}^{\text{3}}}\text{+7x+9)}$

Therefore, $\text{y=uv}$.

$\Rightarrow \dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{du}}{\text{dv}}\text{.v+u}\text{.}\dfrac{\text{du}}{\text{dx}} $ 

$\Rightarrow \dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{d}}{\text{dx}}\text{(}{{\text{x}}^{\text{2}}}\text{-5x+8)}\text{.(}{{\text{x}}^{\text{3}}}\text{+7x+9)+(}{{\text{x}}^{\text{2}}}\text{-5x+8)}\text{.}\dfrac{\text{d}}{\text{dx}}\text{(}{{\text{x}}^{\text{3}}}\text{+7x+9)} $ 

$\Rightarrow \dfrac{\text{dy}}{\text{dx}}\text{=(2x-5)(}{{\text{x}}^{\text{3}}}\text{+7x+9)}\text{.(}{{\text{x}}^{\text{2}}}\text{-5x+8)(3}{{\text{x}}^{\text{2}}}\text{+7)} $ 

$\Rightarrow \dfrac{\text{dy}}{\text{dx}}\text{=2x(}{{\text{x}}^{\text{3}}}\text{+7x+9)-5(}{{\text{x}}^{\text{2}}}\text{-5x+8)+}{{\text{x}}^{\text{2}}}\text{(3}{{\text{x}}^{\text{2}}}\text{+7)-5x(3}{{\text{x}}^{\text{2}}}\text{+7)-8(3}{{\text{x}}^{\text{2}}}\text{+7)} $ 

$\Rightarrow \dfrac{\text{dy}}{\text{dx}}\text{=(2}{{\text{x}}^{\text{4}}}\text{+14}{{\text{x}}^{\text{2}}}\text{+18x)-5}{{\text{x}}^{\text{3}}}\text{-35x-45+(3}{{\text{x}}^{\text{4}}}\text{+7}{{\text{x}}^{\text{2}}}\text{)-15}{{\text{x}}^{\text{3}}}\text{-35x+24}{{\text{x}}^{\text{2}}}\text{+56} $ 

Hence, $\dfrac{\text{dy}}{\text{dx}}\text{=5}{{\text{x}}^{\text{4}}}\text{-20}{{\text{x}}^{\text{3}}}\text{+45}{{\text{x}}^{\text{2}}}\text{+52x+11}$.

(b) By expanding the factors as a polynomial.

Ans.

The given function is

$\text{y=(}{{\text{x}}^{\text{2}}}\text{-5x+8)(}{{\text{x}}^{\text{3}}}\text{+7x+9)}$.

Then, calculating the product, gives

$\text{y=}{{\text{x}}^{\text{2}}}\text{(}{{\text{x}}^{\text{3}}}\text{+7x+9)-5}{{\text{x}}^{\text{4}}}\text{(}{{\text{x}}^{\text{3}}}\text{+7x+9)+8(}{{\text{x}}^{\text{3}}}\text{+7x+9)} $ 

$\Rightarrow \text{y=}{{\text{x}}^{\text{5}}}\text{+7}{{\text{x}}^{\text{3}}}\text{+9}{{\text{x}}^{\text{2}}}\text{-5}{{\text{x}}^{\text{3}}}\text{-26}{{\text{x}}^{\text{2}}}\text{+11x+72} $ 

Now, differentiating both sides of the equation with respect to $\text{x}$ gives $\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{d}}{\text{dx}}\text{(}{{\text{x}}^{\text{5}}}\text{+7}{{\text{x}}^{\text{3}}}\text{+9}{{\text{x}}^{\text{2}}}\text{-5}{{\text{x}}^{\text{3}}}\text{-26}{{\text{x}}^{\text{2}}}\text{+11x+72)} $ 

$=\dfrac{\text{d}}{\text{dx}}\text{(}{{\text{x}}^{\text{5}}}\text{)-5}\dfrac{\text{d}}{\text{dx}}\text{(}{{\text{x}}^{\text{4}}}\text{)+15}\dfrac{\text{d}}{\text{dx}}\text{(}{{\text{x}}^{\text{3}}}\text{)-26}\dfrac{\text{d}}{\text{dx}}\text{(}{{\text{x}}^{\text{3}}}\text{)+11}\dfrac{\text{d}}{\text{dx}}\text{(x)+}\dfrac{\text{d}}{\text{dx}}\text{(72)} $ 

$\text{=5}{{\text{x}}^{\text{4}}}\text{-5 }\!\!\times\!\!\text{ 4}{{\text{x}}^{\text{3}}}\text{+15 }\!\!\times\!\!\text{ 3}{{\text{x}}^{\text{2}}}\text{-26 }\!\!\times\!\!\text{ 2x+11 }\!\!\times\!\!\text{ 1+0} $ 

Hence, $\dfrac{\text{dy}}{\text{dx}}\text{=5}{{\text{x}}^{\text{4}}}\text{-20}{{\text{x}}^{\text{3}}}\text{+45}{{\text{x}}^{\text{2}}}\text{-52x+11}$.

(c) By using a logarithmic function.

Ans.    

The given function is

$\text{y=(}{{\text{x}}^{\text{2}}}\text{-5x+8)(}{{\text{x}}^{\text{3}}}\text{+7x+9)}$.

Now, taking logarithm both sides of the function give

$\text{logy=log(}{{\text{x}}^{\text{2}}}\text{-5x+8)+log(}{{\text{x}}^{\text{3}}}\text{+7x+9)}$

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

$\dfrac{\text{1}}{\text{y}}\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{d}}{\text{dx}}\text{log(}{{\text{x}}^{\text{2}}}\text{-5x+8)+}\dfrac{\text{d}}{\text{dx}}\text{log(}{{\text{x}}^{\text{3}}}\text{+7x+9)} $ 

$\Rightarrow \dfrac{\text{1}}{\text{y}}\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{1}}{{{\text{x}}^{\text{2}}}\text{-5x+8}}\text{.}\dfrac{\text{d}}{\text{dx}}\text{(}{{\text{x}}^{\text{2}}}\text{-5x+8)+}\dfrac{\text{1}}{{{\text{x}}^{\text{3}}}\text{+7x+9}}\text{.}\dfrac{\text{d}}{\text{dx}}\text{(}{{\text{x}}^{\text{3}}}\text{+7x+9)} $ 

$\Rightarrow \dfrac{\text{dy}}{\text{dx}}\text{=y}\left[ \dfrac{\text{1}}{{{\text{x}}^{\text{2}}}\text{-5x+8}}\text{ }\!\!\times\!\!\text{ (2x-5)+}\dfrac{\text{1}}{{{\text{x}}^{\text{3}}}\text{+7x+9}}\text{ }\!\!\times\!\!\text{ (3}{{\text{x}}^{\text{2}}}\text{+7)} \right] $ 

$\Rightarrow \dfrac{\text{dy}}{\text{dx}}\text{=(}{{\text{x}}^{\text{2}}}\text{-5x+8)(}{{\text{x}}^{\text{3}}}\text{+7x+9)}\left[ \dfrac{\text{2x-5}}{{{\text{x}}^{\text{3}}}\text{-5x+8}}\text{+}\dfrac{\text{3}{{\text{x}}^{\text{2}}}\text{+7}}{{{\text{x}}^{\text{3}}}\text{+7x+9}} \right] $ 

$\Rightarrow \dfrac{\text{dy}}{\text{dx}}\text{=(}{{\text{x}}^{\text{2}}}\text{-5x+8)(}{{\text{x}}^{\text{3}}}\text{+7x+9)}\left[ \dfrac{\text{(2x-5)(}{{\text{x}}^{\text{3}}}\text{+7x+9)+(3}{{\text{x}}^{\text{2}}}\text{+7)(}{{\text{x}}^{\text{2}}}\text{-5x+8)}}{\text{(}{{\text{x}}^{\text{3}}}\text{-5x+8)+(}{{\text{x}}^{\text{3}}}\text{+7x+9)}} \right] $ 

$\Rightarrow \dfrac{\text{dy}}{\text{dx}}\text{=2x(}{{\text{x}}^{\text{3}}}\text{+7x+9}{{\text{x}}^{\text{2}}}\text{)-5(}{{\text{x}}^{\text{3}}}\text{+7x+9)+3}{{\text{x}}^{\text{2}}}\text{(}{{\text{x}}^{\text{2}}}\text{-5x+8)+7(}{{\text{x}}^{\text{3}}}\text{+7x+9)} $ 

$\Rightarrow \dfrac{\text{dy}}{\text{dx}}\text{=(2}{{\text{x}}^{\text{4}}}\text{+14}{{\text{x}}^{\text{2}}}\text{+18x)+(5}{{\text{x}}^{\text{3}}}\text{-35x+45)+(3}{{\text{x}}^{\text{4}}}\text{-15}{{\text{x}}^{\text{3}}}\text{+24}{{\text{x}}^{\text{2}}}\text{)+(7}{{\text{x}}^{\text{2}}}\text{+35x+56)} $ 

Therefore, $\dfrac{\text{dy}}{\text{dx}}\text{=5}{{\text{x}}^{\text{2}}}\text{-20}{{\text{x}}^{\text{3}}}\text{+45}{{\text{x}}^{\text{2}}}\text{-52x+11}$.

Hence, comparing the above three results, it is concluded that the derivative $\dfrac{\text{dy}}{\text{dx}}$ are the same for all methods.

18. Let $\mathbf{u}$, $\mathbf{v}$, and $\mathbf{w}$ are functions of $\mathbf{x}$ , then prove that $\dfrac{\text{d}}{\text{dx}}\text{(u}\text{.v}\text{.w)=}\dfrac{\text{du}}{\text{dx}}\text{v}\text{.w+u}\dfrac{\text{du}}{\text{dx}}\text{.w+u}\text{.v}\dfrac{\text{dw}}{\text{d}\mathbf{x}}$ in two ways. First by using repeated application of product rule and second by applying logarithmic differentiation. 

Ans.

Let the function $\text{y=u}\text{.v}\text{.w=u}\text{.(v}\text{.w)}$.

Then applying the product rule of derivatives, give

$\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{du}}{\text{dx}}\text{.(v}\text{.w)+u}\text{.}\dfrac{\text{d}}{\text{dx}}\text{(v}\text{.w)}$          

$\Rightarrow \dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{du}}{\text{dx}}\text{v}\text{.w+u}\left[ \dfrac{\text{dv}}{\text{dx}}\text{.w+v}\text{.}\dfrac{\text{dv}}{\text{dx}} \right]$        (Using the product rule again)

Thus,

$\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{du}}{\text{dx}}\text{v}\text{.w+u}\text{.}\dfrac{\text{dv}}{\text{dx}}\text{.w+u}\text{.v}\dfrac{\text{dw}}{\text{dx}}$.

Now, take the logarithm of both sides of the function $\text{y=u}\text{.v}\text{.w}$.

Then, we have $\text{logy=logu+logv+logw}$.

Differentiating both sides of the equation with respect to $\text{x}$ gives $\dfrac{\text{1}}{\text{y}}\text{.}\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{d}}{\text{dx}}\text{(logu)+}\dfrac{\text{d}}{\text{dx}}\text{(logv)+}\dfrac{\text{d}}{\text{dx}}\text{(logw)} $ 

$\Rightarrow \dfrac{\text{1}}{\text{y}}\text{.}\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{1}}{\text{u}}\dfrac{\text{du}}{\text{dx}}\text{+}\dfrac{\text{1}}{\text{v}}\dfrac{\text{dv}}{\text{dx}}\text{+}\dfrac{\text{1}}{\text{w}}\dfrac{\text{dw}}{\text{dx}} $ 

$\Rightarrow \dfrac{\text{dy}}{\text{dx}}\text{=y}\left( \dfrac{\text{1}}{\text{u}}\dfrac{\text{du}}{\text{dx}}\text{+}\dfrac{\text{1}}{\text{v}}\dfrac{\text{dv}}{\text{dx}}\text{+}\dfrac{\text{1}}{\text{w}}\dfrac{\text{dw}}{\text{dx}} \right) $ 

$\Rightarrow \dfrac{\text{dy}}{\text{dx}}\text{=u}\text{.v}\text{.w}\left( \dfrac{\text{1}}{\text{u}}\dfrac{\text{du}}{\text{dx}}\text{+}\dfrac{\text{1}}{\text{v}}\dfrac{\text{dv}}{\text{dx}}\text{+}\dfrac{\text{1}}{\text{w}}\dfrac{\text{dw}}{\text{dx}} \right) $ 

Hence, $\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{du}}{\text{dx}}\text{v}\text{.w+u}\dfrac{\text{du}}{\text{dx}}\text{.w+u}\text{.v}\dfrac{\text{dw}}{\text{dx}}$.

Conclusion

Class 12 Maths Chapter 5 Exercise 5.5 - Continuity and Differentiability, is crucial for a solid foundation in math. Understanding the concepts of differentiability and continuity of functions. Referring to these NCERT Solutions by Vedantu can significantly enhance your understanding of continuity and differentiability. Regular practice with Class 12 Exercise 5.5 NCERT solutions provided by platforms like Vedantu can enhance comprehension and problem-solving skills. Pay attention to the step-by-step solutions provided, grasp the underlying principles, and ensure clarity on the concepts before moving forward.

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