Chapter 5: Continuity and Differentiability

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

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

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

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

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

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

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

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

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

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.

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