Chapter 5: Continuity and Differentiability

Ans:

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

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

$\frac{\text{dy}}{\text{dx}}\text{=}\frac{\text{d}}{\text{dx}}\text{(}{{\text{x}}^{\text{2}}}\text{)+}\frac{\text{d}}{\text{dx}}\text{(3x)+}\frac{\text{d}}{\text{dx}}\text{(2)=2x+3+0=2x+3}$

That is,

$\frac{\text{dy}}{\text{dx}}=2\text{x+3}$.

Again, differentiating both sides with respect to $\text{x}$ gives

$\frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{=}\frac{\text{d}}{\text{dx}}\text{(2x+3)=}\frac{\text{d}}{\text{dx}}\text{(2x)+}\frac{\text{d}}{\text{dx}}\text{(3)=2+0=2}$

Hence, $\frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{=2}$.

Ans: 

The given function is $\text{y=}{{\text{x}}^{\text{20}}}$.

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

\[\frac{\text{dy}}{\text{dx}}\text{=}\frac{\text{d}}{\text{dx}}\text{(}{{\text{x}}^{\text{20}}}\text{)=20}{{\text{x}}^{\text{19}}}\]

Again, differentiating both sides with respect to $\text{x}$ gives

\[\frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{=}\frac{\text{d}}{\text{dx}}\text{(20}{{\text{x}}^{\text{19}}}\text{)=20}\frac{\text{d}}{\text{dx}}\text{(}{{\text{x}}^{\text{19}}}\text{)=20}\left( 19 \right){{\text{x}}^{\text{18}}}\text{=380}{{\text{x}}^{\text{18}}}\].

Hence, \[\frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{=380}{{\text{x}}^{\text{18}}}\].

Ans: 

The given function is $\text{y=x}\text{.cosx}$.

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

$\frac{\text{dy}}{\text{dx}}\text{=}\frac{\text{d}}{\text{dx}}\text{(x}\text{.cosx)=cosx}\text{.}\frac{\text{d}}{\text{dx}}\text{(x)+x}\frac{\text{d}}{\text{dx}}\text{(cosx)=cosx}\text{.1+x(-sinx)=cosx-xsinx}$

That is, $\frac{\text{dy}}{\text{dx}}\text{=cosx-xsinx}$.

Again, differentiating both sides with respect to $\text{x}$ gives

$\begin{align}

  & \frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{=}\frac{\text{d}}{\text{dx}}\text{(cosx-xsinx)=}\frac{\text{d}}{\text{dx}}\text{(cosx)-}\frac{\text{d}}{\text{dx}}\text{(xsinx)} \\ 

 & \text{=-sinx- }\!\![\!\!\text{ sinx}\cdot \frac{\text{d}}{\text{dx}}\text{(x)+x}\cdot \frac{\text{d}}{\text{dx}}\text{(sinx) }\!\!]\!\!\text{ } \\ 

 & \text{=-sinx-(sinx+xcosx)} \\ 

\end{align}$

Hence, $\frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{=-(xcosx+2sinx)}$.

Ans: 

The given function is $\text{y=logx}$.

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

\[\frac{\text{dy}}{\text{dx}}\text{=}\frac{\text{d}}{\text{dx}}\text{(logx)=}\frac{\text{1}}{\text{x}}\]

Again, differentiating both sides with respect to $\text{x}$ gives

\[\frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{=}\frac{\text{d}}{\text{dx}}\left( \frac{\text{1}}{\text{x}} \right)\text{=}\frac{\text{-1}}{{{\text{x}}^{\text{2}}}}\]

Hence, \[\frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{=-}\frac{\text{1}}{{{\text{x}}^{\text{2}}}}\].

Ans: 

The given function is $\text{y=}{{\text{x}}^{\text{3}}}\text{logx}$.

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

$\frac{\text{dy}}{\text{dx}}\text{=}\frac{\text{d}}{\text{dx}}\left[ {{\text{x}}^{\text{3}}}\text{logx} \right]\text{=logx}\text{.}\frac{\text{d}}{\text{dx}}\text{(}{{\text{x}}^{\text{3}}}\text{)+}{{\text{x}}^{\text{3}}}\frac{\text{d}}{\text{dx}}\text{(logx)=logx}\text{.3}{{\text{x}}^{\text{2}}}\text{+}{{\text{x}}^{\text{3}}}\text{.}\frac{\text{1}}{\text{x}}\text{=logx}\text{.3}{{\text{x}}^{\text{2}}}\text{+}{{\text{x}}^{\text{2}}}$

That is, $\frac{\text{dy}}{\text{dx}}\text{=}{{\text{x}}^{\text{2}}}\text{(1+3logx)}$. 

Again, differentiating both sides with respect to $\text{x}$ gives

$\begin{align}

  & \frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{=}\frac{\text{d}}{\text{dx}}\text{(}{{\text{x}}^{\text{2}}}\text{(1+3logx))} \\ 

 & \text{=(1+3logx)}\text{.}\frac{\text{d}}{\text{dx}}\text{(}{{\text{x}}^{\text{2}}}\text{)+}{{\text{x}}^{\text{2}}}\frac{\text{d}}{\text{dx}}\text{(1+3logx)} \\ 

 & \text{=(1+3logx)}\text{.2x+}{{\text{x}}^{\text{3}}}\text{.}\frac{\text{3}}{\text{x}} \\ 

 & \text{=2x+6logx+3x} \\ 

 & \text{=5x+6xlogx}  

\end{align}$

Hence, $\frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{=x(5+6logx)}$.

Ans: 

The given function is $\text{y=}{{\text{e}}^{\text{x}}}\text{sin5x}$.

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

\[\begin{align}

  & \frac{\text{dy}}{\text{dx}}\text{=}\frac{\text{d}}{\text{dx}}\left[ {{\text{e}}^{\text{x}}}\text{sin5x} \right]\text{=sinx}\frac{\text{d}}{\text{dx}}\text{(}{{\text{e}}^{\text{x}}}\text{)+}{{\text{e}}^{\text{x}}}\frac{\text{d}}{\text{dx}}\text{(sin5x)} \\ 

 & \Rightarrow \frac{\text{dy}}{\text{dx}}\text{=sin5x}\text{.}{{\text{e}}^{\text{x}}}\text{+}{{\text{e}}^{\text{x}}}\text{.cos5x}\text{.}\frac{\text{d}}{\text{dx}}\text{(5x)} \\ 

\end{align}\]

That is, $\frac{\text{dy}}{\text{dx}}\text{=}{{\text{e}}^{\text{x}}}\text{(sin5x+5cos5x)}$.

Again, differentiating both sides with respect to $\text{x}$ gives

\[\begin{align}

  & \frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{=}\frac{\text{d}}{\text{dx}}\left[ {{\text{e}}^{\text{x}}}\text{(sin5x+5cos5x)} \right] \\ 

 & \text{=(sin5x+5cos5x)}\text{.}\frac{\text{d}}{\text{dx}}\text{(}{{\text{e}}^{\text{x}}}\text{)+}{{\text{e}}^{\text{x}}}\text{.}\frac{\text{d}}{\text{dx}}\text{(sin5x+5cos5x)} \\ 

 & \text{=(sin5x+5cos5x)(}{{\text{e}}^{\text{x}}}\text{)+}{{\text{e}}^{\text{x}}}\left[ \text{cos5x}\text{.}\frac{\text{d}}{\text{dx}}\text{(5x)+5(-sin5x)}\text{.}\frac{\text{d}}{\text{dx}}\text{(5x)} \right] \\ 

 & \text{=}{{\text{e}}^{\text{x}}}\text{(sin5x+5cos5x)+}{{\text{e}}^{\text{x}}}\text{(5cos5x-25sin5x)} \\ 

\end{align}\]

 $\text{=}{{\text{e}}^{\text{x}}}\text{(10cos5x-24sin5x)}$.

Hence, $\frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{=2}{{\text{e}}^{\text{x}}}\text{(5cox5x-12sin5x)}$.

Ans:  

The given function is $\text{y=}{{\text{e}}^{\text{6x}}}\text{cos3x}$.

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

$\begin{align}

  & \frac{\text{dy}}{\text{dx}}\text{=}\frac{\text{d}}{\text{dx}}\text{(}{{\text{e}}^{\text{6x}}}\text{cos3x)=cos3x }\!\!\times\!\!\text{ }\frac{\text{d}}{\text{dx}}\text{(}{{\text{e}}^{\text{6x}}}\text{)+}{{\text{e}}^{\text{6x}}}\text{ }\!\!\times\!\!\text{ }\frac{\text{d}}{\text{dx}}\text{(cos3x)} \\ 

 & \Rightarrow \frac{\text{dy}}{\text{dx}}=\text{cos3x }\!\!\times\!\!\text{ }{{\text{e}}^{\text{6x}}}\text{ }\!\!\times\!\!\text{ }\frac{\text{d}}{\text{dx}}\text{(6x)+}{{\text{e}}^{\text{6x}}}\text{ }\!\!\times\!\!\text{ (-sin3x) }\!\!\times\!\!\text{ }\frac{\text{d}}{\text{dx}}\text{(3x)} \\ 

\end{align}$

Therefore,

$\frac{\text{dy}}{\text{dx}}\text{=6}{{\text{e}}^{\text{6x}}}\text{cos3x-3}{{\text{e}}^{\text{6x}}}\text{sin3x}$ …… (1)

Again, differentiating both sides with respect to $\text{x}$ gives

$\frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{=}\frac{\text{d}}{\text{dx}}\text{(6}{{\text{e}}^{\text{6x}}}\text{cos3x-3}{{\text{e}}^{\text{6x}}}\text{sin3x)=6 }\!\!\times\!\!\text{ }\frac{\text{d}}{\text{dx}}\text{(}{{\text{e}}^{\text{6x}}}\text{cos3x)-3 }\!\!\times\!\!\text{ }\frac{\text{d}}{\text{dx}}\text{(}{{\text{e}}^{\text{6x}}}\text{sin3x)}$

$\text{=6 }\!\!\times\!\!\text{ }\left[ \text{6}{{\text{e}}^{\text{6x}}}\text{cos3x-3}{{\text{e}}^{\text{6x}}}\text{sin3x} \right]\text{-3 }\!\!\times\!\!\text{ }\left[ \text{sin3x }\!\!\times\!\!\text{ }\frac{\text{d}}{\text{dx}}\text{(}{{\text{e}}^{\text{6x}}}\text{)+}{{\text{e}}^{\text{6x}}}\text{ }\!\!\times\!\!\text{ }\frac{\text{d}}{\text{dx}}\text{(sin3x)} \right]$ [using (1)]

\[\begin{align}

  & \text{=36}{{\text{e}}^{\text{6x}}}\text{cos3x-18}{{\text{e}}^{\text{6x}}}\text{sin3x-3}\left[ \text{sin3x }\!\!\times\!\!\text{ }{{\text{e}}^{\text{6x}}}\text{ }\!\!\times\!\!\text{ 6+}{{\text{e}}^{\text{6x}}}\text{ }\!\!\times\!\!\text{ cos3x-3} \right] \\ 

 & \text{=36}{{\text{e}}^{\text{6x}}}\text{cos3x-18}{{\text{e}}^{\text{6x}}}\text{sin3x-18}{{\text{e}}^{\text{6x}}}\text{sin3x-9}{{\text{e}}^{\text{6x}}}\text{cos3x} \\ 

\end{align}\]

Hence, \[\frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{=9}{{\text{e}}^{\text{6x}}}\text{(3cos3x-4sin3x)}\].

Ans: 

The given function is $\text{y=ta}{{\text{n}}^{\text{-1}}}\text{x}$.

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

$\frac{\text{dy}}{\text{dx}}\text{=}\frac{\text{d}}{\text{dx}}\text{ta}{{\text{n}}^{\text{-1}}}\text{x=}\frac{\text{1}}{\text{1-}{{\text{x}}^{\text{2}}}}$

Again, differentiating both sides with respect to $\text{x}$ gives

$\begin{align}

  & \frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{=}\frac{\text{d}}{\text{dx}}\left( \frac{\text{1}}{\text{1+}{{\text{x}}^{\text{2}}}} \right)\text{=}\frac{\text{d}}{\text{dx}}{{\text{(1+}{{\text{x}}^{\text{2}}}\text{)}}^{\text{-1}}}\text{=(-1) }\!\!\times\!\!\text{ (1+}{{\text{x}}^{\text{2}}}{{\text{)}}^{\mathbf{-2}}}\text{ }\!\!\times\!\!\text{ }\frac{\text{d}}{\text{dx}}\text{(1+}{{\text{x}}^{\text{2}}}\text{)} \\ 

 & \text{=-}\frac{\text{1}}{{{\text{(1+}{{\text{x}}^{\text{2}}}\text{)}}^{2}}}\text{ }\!\!\times\!\!\text{ 2x} \\ 

\end{align}$

Hence, $\frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{=}\frac{\text{-2x}}{{{\text{(1+}{{\text{x}}^{\text{2}}}\text{)}}^{2}}}$.

Ans: 

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

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

$\begin{align}

  & \frac{\text{dy}}{\text{dx}}\text{=}\frac{\text{d}}{\text{dx}}\text{ }\!\![\!\!\text{ log(logx) }\!\!]\!\!\text{ } \\ 

 & =\frac{\text{1}}{\text{logx}}\text{ }\!\!\times\!\!\text{ }\frac{\text{d}}{\text{dx}}\text{(logx)} \\ 

 & \Rightarrow \frac{\text{dy}}{\text{dx}}\text{=(xlogx}{{\text{)}}^{\text{-1}}} \\ 

\end{align}$

Again, differentiating both sides with respect to $\text{x}$ gives

$\begin{align}

  & \frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{=}\frac{\text{d}}{\text{dx}}\left[ {{\text{(xlogx)}}^{\text{-1}}} \right]\text{=(-1) }\!\!\times\!\!\text{ (xlogx}{{\text{)}}^{\text{-2}}}\frac{\text{d}}{\text{dx}}\text{(xlogx)} \\ 

 & =\frac{\text{-1}}{{{\text{(xlogx)}}^{\text{2}}}}\text{ }\!\!\times\!\!\text{ }\left[ \text{logx }\!\!\times\!\!\text{ }\frac{\text{d}}{\text{dx}}\text{(x)+x }\!\!\times\!\!\text{ }\frac{\text{d}}{\text{dx}}\text{(logx)} \right] \\ 

 & =\frac{\text{-1}}{{{\text{(xlogx)}}^{\text{2}}}}\text{ }\!\!\times\!\!\text{ }\left[ \text{logx }\!\!\times\!\!\text{ 1+x }\!\!\times\!\!\text{ }\frac{\text{1}}{\text{x}} \right] \\ 

\end{align}$

Hence, $\frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{=}\frac{\text{-(1+logx)}}{{{\text{(xlogx)}}^{\text{2}}}}$.

Ans: 

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

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

\[\frac{\text{dy}}{\text{dx}}\text{=}\frac{\text{d}}{\text{dx}}\text{ }\!\![\!\!\text{ sin(logx) }\!\!]\!\!\text{ =cos(logx) }\!\!\times\!\!\text{ }\frac{\text{d}}{\text{dx}}\text{(logx)}=\frac{\text{cos(logx)}}{\text{x}}\]

Again, differentiating both sides with respect to $\text{x}$ gives

\[\begin{align}

  & \frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{=}\frac{\text{d}}{\text{dx}}\left[ \frac{\text{cos(logx)}}{\text{x}} \right] \\ 

 & =\frac{\text{x}\left[ \text{cos(logx)} \right]\text{-cos(logx) }\!\!\times\!\!\text{ }\frac{\text{d}}{\text{dx}}\text{(x)}}{{{\text{x}}^{\text{2}}}} \\ 

 & =\frac{\text{x}\left[ \text{-sin(logx) }\!\!\times\!\!\text{ }\frac{\text{d}}{\text{dx}}\text{(logx)} \right]\text{-cos(logx) }\!\!\times\!\!\text{ 1}}{{{\text{x}}^{\text{2}}}} \\ 

 & =\frac{\text{-xsin(logx) }\!\!\times\!\!\text{ }\frac{\text{1}}{\text{x}}\text{-cos(logx)}}{{{\text{x}}^{\text{2}}}} \\ 

\end{align}\]

Hence, \[\frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{=}\frac{\left[ \text{-sin(logx)-cos(logx)} \right]}{{{\text{x}}^{\text{2}}}}\].