Chapter 5: Continuity and Differentiability

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,

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

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,

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

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,

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

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

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

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,

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

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

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

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

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

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

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

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

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

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

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