NCERT Solutions for Class 12 Maths Chapter 5: Continuity and Differentiability - Exercise 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, $\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{du}}{\text{dx}}\text{v}\text{.w+u}\dfrac{\text{du}}{\text{dx}}\text{.w+u}\text{.v}\dfrac{\text{dw}}{\text{dx}}$.

Class 12 Continuity and Differentiability Exercise 5.5

NCERT Solutions for Class 12 Math Chapter 5 Exercise 5.5. PDF given on this page consists of solutions for all the questions given in the exercise. These NCERT solutions are based on the latest Class 12 CBSE guidelines and are extremely helpful for quick revisions. Download the NCERT Solutions of Class 12 Math Chapter 5 Continuity and Differentiability Ex-5.5 and practise them for your exam preparations.

Topics Covered in NCERT Class 12 Maths Chapter 5 Exercise 5.5

Class 12 Maths Chapter 5 Exercise 5.5 covers sums based on the topic of Logarithmic Differentiation. Logarithm differentiation is a method of finding the derivative of a function. In this method, the derivative of a function is calculated simply by first finding the logarithm of a given function and then differentiating it.

The logarithm differentiation method is generally used when the value of a function is given in the form of y = f(x)g(x).

What are the Steps to Solve Logarithmic Differentiation?

The steps to calculate logarithm differentiation are discussed below.

1. Take log on both sides

2. Use log property to remove exponent

3. Now differentiate the equation

4. Simplify the obtained equation

5. Substitute back the value of y in the equation

Summary of NCERT Solution of Maths Class 12 Chapter 5 Exercise 5.5

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NCERT Solutions for Class 12 Maths

• Chapter 1 - Relations and Functions

• Chapter 2 - Inverse Trigonometric Functions

• Chapter 3 - Matrices

• Chapter 4 - Determinants

• Chapter 5 - Continuity and Differentiability

• Chapter 6 - Application of Derivatives

• Chapter 7 - Integrals

• Chapter 8 - Application of Integrals

• Chapter 9 - Differential Equations

• Chapter 10 - Vector Algebra

• Chapter 11 - Three Dimensional Geometry

• Chapter 12 - Linear Programming

• Chapter 13 - Probability

Class 12 Maths Chapter 5 Exercises