Chapter 5: Continuity and Differentiability

Ans: 

The given equation is $\text{y=5cosx-3sinx}$.

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

$\frac{\text{dy}}{\text{dx}}\text{=}\frac{\text{d}}{\text{dx}}\text{(5cosx)-}\frac{\text{d}}{\text{dx}}\text{(3sinx)=5}\frac{\text{d}}{\text{dx}}\text{(cosx)-3}\frac{\text{d}}{\text{dx}}\text{(sinx)=5(-sinx)-3cosx}$

Therefore, $\frac{\text{dy}}{\text{dx}}\text{=-(5sinx+3cosx)}$.

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

$\frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{=}\frac{\text{d}}{\text{dx}}\text{ }\!\![\!\!\text{ -(5sinx+3cosx) }\!\!]\!\!\text{ }$

$\begin{align}

  & \text{=-}\left[ \text{5 }\!\!\times\!\!\text{ }\frac{\text{d}}{\text{dx}}\text{(sinx)+3 }\!\!\times\!\!\text{ }\frac{\text{d}}{\text{dx}}\text{(cosx)} \right] \\ 

 & \text{= }\!\![\!\!\text{ 5cosx+3(-sinx) }\!\!]\!\!\text{ } \\ 

 & \text{=-y} \\ 

\end{align}$

That is, $\frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{=-y}$.

Hence, $\frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{+y=0}$.

Ans: 

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

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

$\frac{\text{dy}}{\text{dx}}\text{=}\frac{\text{d}}{\text{dx}}\text{(co}{{\text{s}}^{\text{-1}}}\text{x)=}\frac{\text{-1}}{\sqrt{\text{1-}{{\text{x}}^{\text{2}}}}}\text{=-(1-}{{\text{x}}^{\text{2}}}{{\text{)}}^{\frac{\text{-1}}{\text{2}}}}$

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[ \text{-(1-}{{\text{x}}^{\text{2}}}{{\text{)}}^{\frac{\text{-1}}{\text{2}}}} \right] \\ 

 & =\left( \frac{\text{-1}}{\text{2}} \right)\text{ }\!\!\times\!\!\text{ (1-}{{\text{x}}^{\text{2}}}{{\text{)}}^{\frac{\text{-3}}{\text{2}}}}\text{ }\!\!\times\!\!\text{ }\frac{\text{d}}{\text{dx}}\text{(1-}{{\text{x}}^{\text{2}}}\text{)} \\ 

 & =\frac{\text{1}}{\sqrt[\text{2}]{{{\text{(1-}{{\text{x}}^{\text{2}}}\text{)}}^{\text{3}}}}}\text{ }\!\!\times\!\!\text{ (-2x)} \\ 

\end{align}$

$\Rightarrow \frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{=}\frac{\text{-x}}{\sqrt{{{\text{(1-}{{\text{x}}^{\text{2}}}\text{)}}^{\text{3}}}}}$                                                                       …… (1)

Now, $\text{y=co}{{\text{s}}^{\text{-1}}}\text{x}\Rightarrow \text{x=cosy}$.

Therefore, substituting $\text{x=cosy}$ into equation (1), gives

$\begin{align}

  & \frac{{{\text{d}}^{\text{2}}}\text{x}}{\text{d}{{\text{y}}^{\text{2}}}}\text{=}\frac{\text{-cosy}}{\sqrt{{{\text{(1-co}{{\text{s}}^{\text{2}}}\text{y)}}^{\text{3}}}}} \\ 

 & =\frac{\text{-cosy}}{\text{si}{{\text{n}}^{\text{3}}}\text{y}} \\ 

 & =\frac{\text{-cosy}}{\text{siny}}\text{ }\!\!\times\!\!\text{ }\frac{\text{1}}{\text{si}{{\text{n}}^{\text{2}}}\text{y}} \\ 

\end{align}$

Hence, $\frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{= -coty }\!\!\times\!\!\text{ cose}{{\text{c}}^{\text{2}}}\text{y}$.

Ans: 

The given equations are $\text{y=3cos(logx)+4sin(logx)}$                               …… (1)

and ${{\text{x}}^{\text{2}}}{{\text{y}}_{\text{2}}}\text{+x}{{\text{y}}_{\text{1}}}\text{+y=0}$                                                                                …… (2)

Then, differentiating both sides of the equation (1) with respect to $\text{x}$ gives

\[\begin{align}

  & {{\text{y}}_{\text{1}}}\text{=3 }\!\!\times\!\!\text{ }\frac{\text{d}}{\text{dx}}\text{ }\!\![\!\!\text{ cos(logx) }\!\!]\!\!\text{ +4 }\!\!\times\!\!\text{ }\frac{\text{d}}{\text{dx}}\text{ }\!\![\!\!\text{ sin(logx) }\!\!]\!\!\text{ } \\ 

 & \text{=3 }\!\!\times\!\!\text{ }\left[ \text{-sin(logx) }\!\!\times\!\!\text{ }\frac{\text{d}}{\text{dx}}\text{(logx)} \right]\text{+4 }\!\!\times\!\!\text{ }\left[ \text{cos(logx) }\!\!\times\!\!\text{ }\frac{\text{d}}{\text{dx}}\text{(logx)} \right] \\ 

\end{align}\]

\[{{\text{y}}_{\text{1}}}\text{=}\frac{\text{-3sin(logx)}}{\text{x}}\text{+}\frac{\text{4cos(logx)}}{\text{x}}\text{=}\frac{\text{4cos(logx)-3sin(logx)}}{\text{x}}\]

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

\[{{\text{y}}_{\text{2}}}\text{=}\frac{\text{d}}{\text{dx}}\left( \frac{\text{4cos(logx)-3sin(logx)}}{\text{x}} \right)\]

\[\begin{align}

  & \text{=}\frac{\text{x}\cdot \frac{d}{dx}\left[ \text{4 }\!\!\{\!\!\text{ cos(logx) }\!\!\}\!\!\text{ - }\!\!\{\!\!\text{ 3sin(logx) }\!\!\}\!\!\text{ } \right]\text{- }\!\!\{\!\!\text{ 4cos(logx)-3sin(logx) }\!\!\}\!\!\text{ }\times \text{1}}{{{\text{x}}^{\text{2}}}} \\ 

 & \text{=}\frac{\text{x}\left[ \text{-4sin(logx)}\frac{d}{dx}\text{(logx)-3cos(logx)}\frac{d}{dx}\text{(logx)} \right]\text{-4cos(logx)+3sin(logx)}}{{{\text{x}}^{\text{2}}}} \\ 

\end{align}\]

$=\dfrac{\text{x}\left[ \text{-4sin(logx)}\cdot \dfrac{\text{1}}{\text{x}}\text{-3cos(logx)}\cdot \dfrac{\text{1}}{\text{x}} \right]\text{-4cos(logx)+3sin(logx)}}{{{\text{x}}^{\text{2}}}}$

$=\dfrac{\text{-4sin(logx)-3cos(logx)-4cos(logx)+3sin(logx)}}{{{\text{x}}^{\text{2}}}}$

Therefore, ${{\text{y}}_{\text{2}}}\text{=}\dfrac{\text{-sin(logx)-7cos(logx)}}{{{\text{x}}^{\text{2}}}}$.

Now, substituting the derivatives ${{\text{y}}_{1}}$ ,${{\text{y}}_{2}}$ and $\text{y}$ into the LHS of the equation (2) gives

$\begin{align}

  & {{\text{x}}^{\text{2}}}{{\text{y}}_{\text{2}}}\text{+x}{{\text{y}}_{\text{1}}}\text{+y} \\ 

 & \text{=}{{\text{x}}^{\text{2}}}\left( \frac{\text{-sin(logx)-7cos(logx)}}{{{\text{x}}^{\text{2}}}} \right)\text{+x}\left( \frac{\text{4cos(logx)-3sin(logx)}}{{{\text{x}}^{\text{2}}}} \right)\text{+3cos(logx)+4sin(logx)} \\ 

 & \text{=-sin(logx)-7cos(logx)+4cos(logx)-3sin(logx)+4sin(logx)} \\ 

 & \text{=0} \\ 

\end{align}$

Hence, it has been proved that ${{\text{x}}^{\text{2}}}{{\text{y}}_{\text{2}}}\text{+x}{{\text{y}}_{\text{1}}}\text{+y}=0$.

Ans: 

The given equations are $\text{y=A}{{\text{e}}^{\text{mx}}}\text{+B}{{\text{e}}^{\text{nx}}}$                                     …… (1)

and \[\frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{-(m+n)}\frac{\text{dy}}{\text{dx}}\text{+mny=0}\]                                                      ……. (2)

Then, differentiating both sides of the equation (1) with respect to $\text{x}$ gives

$\frac{\text{dy}}{\text{dx}}\text{=A}\text{.}\frac{\text{d}}{\text{dx}}\text{(}{{\text{e}}^{\text{mx}}}\text{)+B}\text{.}\frac{\text{d}}{\text{dx}}\text{(}{{\text{e}}^{\text{mx}}}\text{)=A}\text{.}{{\text{e}}^{\text{mx}}}\text{.}\frac{\text{d}}{\text{dx}}\text{(mx)+B}\text{.}{{\text{e}}^{\text{nx}}}\text{.}\frac{\text{d}}{\text{dx}}\text{(nx)=Am}{{\text{e}}^{\text{mx}}}\text{+Bn}{{\text{e}}^{\text{nx}}}$

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{(Am}{{\text{e}}^{\text{mx}}}\text{+Bn}{{\text{e}}^{\text{nx}}}\text{)=Am}\text{.}\frac{\text{d}}{\text{dx}}\text{(}{{\text{e}}^{\text{mx}}}\text{)+Bn}\text{.}\frac{\text{d}}{\text{dx}}\text{(}{{\text{e}}^{\text{nx}}}\text{)} \\ 

 & \text{=Am}\text{.}{{\text{e}}^{\text{mx}}}\text{.}\frac{\text{d}}{\text{dx}}\text{(mx)+Bn}\text{.}{{\text{e}}^{\text{nx}}}\text{.}\frac{\text{d}}{\text{dx}}\text{(nx)} \\ 

\end{align}$

Therefore, $\frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{=A}{{\text{m}}^{\text{2}}}{{\text{e}}^{\text{mx}}}\text{+B}{{\text{n}}^{\text{2}}}{{\text{e}}^{\text{nx}}}$.

Thus, substituting the derivatives ${{\text{y}}_{1}}$ ,${{\text{y}}_{2}}$ and $\text{y}$ into the LHS of the equation (2) gives

$\begin{align}

  & \frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{-(m+n)}\frac{\text{dy}}{\text{dx}}\text{+mny} \\ 

 & \text{=A}{{\text{m}}^{\text{2}}}\text{e}{{\text{x}}^{\text{mx}}}\text{+B}{{\text{n}}^{\text{2}}}{{\text{e}}^{\text{nx}}}\text{-(m+n)}\text{.(Am}{{\text{e}}^{\text{mx}}}\text{+Bn}{{\text{e}}^{\text{nx}}}\text{)+mn(A}{{\text{e}}^{\text{mx}}}\text{+B}{{\text{e}}^{\text{nx}}}\text{)} \\ 

 & \text{=A}{{\text{m}}^{\text{2}}}\text{e}{{\text{x}}^{\text{mx}}}\text{+B}{{\text{n}}^{\text{2}}}{{\text{e}}^{\text{nx}}}\text{-Ame}{{\text{x}}^{\text{mx}}}\text{+Bmn}{{\text{e}}^{\text{nx}}}\text{+Amn}{{\text{e}}^{\text{mx}}}\text{+B}{{\text{n}}^{\text{2}}}{{\text{e}}^{\text{nx}}}\text{+Amn}{{\text{e}}^{\text{mx}}}\text{+Bmn}{{\text{e}}^{\text{nx}}} \\ 

 & \text{=0} \\ 

\end{align}$

Thus, it has been proved that \[\frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{-(m+n)}\frac{\text{dy}}{\text{dx}}\text{+mny=0}\].

Ans: 

The given equation is $\text{y=500}{{\text{e}}^{\text{7x}}}\text{+600}{{\text{e}}^{\text{-7x}}}$.                                     …… (1)

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

$\begin{align}

  & \frac{\text{dy}}{\text{dx}}\text{=500 }\!\!\times\!\!\text{ (}{{\text{e}}^{\text{7x}}}\text{)+600 }\!\!\times\!\!\text{ }\frac{\text{d}}{\text{dx}}\text{(-7x)} \\ 

 & \text{=500 }\!\!\times\!\!\text{ }{{\text{e}}^{\text{7x}}}\text{ }\!\!\times\!\!\text{ }\frac{\text{d}}{\text{dx}}\text{(7x)+600 }\!\!\times\!\!\text{ }{{\text{e}}^{\text{-7x}}}\text{ }\!\!\times\!\!\text{ }\frac{\text{d}}{\text{dx}}\text{(-7x)} \\ 

 & \text{=3500}{{\text{e}}^{\text{7x}}}\text{-4200}{{\text{e}}^{\text{-7x}}} \\ 

\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{=3500 }\!\!\times\!\!\text{ }\frac{\text{d}}{\text{dx}}\text{(}{{\text{e}}^{\text{7x}}}\text{)-4200 }\!\!\times\!\!\text{ }\frac{\text{d}}{\text{dx}}\text{(}{{\text{e}}^{\text{-7x}}}\text{)} \\ 

 & \text{=3500 }\!\!\times\!\!\text{ }{{\text{e}}^{\text{7x}}}\text{ }\!\!\times\!\!\text{ }\frac{\text{d}}{\text{dx}}\text{(7x)-4200 }\!\!\times\!\!\text{ }{{\text{e}}^{\text{-7x}}}\text{ }\!\!\times\!\!\text{ }\frac{\text{d}}{\text{dx}}\text{(-7x)} \\ 

 & \text{=7 }\!\!\times\!\!\text{ 3500 }\!\!\times\!\!\text{ }{{\text{e}}^{\text{7x}}}\text{+7 }\!\!\times\!\!\text{ 4200 }\!\!\times\!\!\text{ }{{\text{e}}^{\text{-7x}}} \\ 

 & \text{=49 }\!\!\times\!\!\text{ 500}{{\text{e}}^{\text{7x}}}\text{+49 }\!\!\times\!\!\text{ 600}{{\text{e}}^{\text{-7x}}} \\ 

 & \text{=49(500}{{\text{e}}^{\text{7x}}}\text{+600}{{\text{e}}^{\text{-7x}}}\text{)} \\ 

\end{align}$

$\text{=49y}$, using the equation (1).

Thus, it has been proved that $\frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{=49y}$.

Ans: 

The given equation is ${{\text{e}}^{\text{y}}}\text{(x+1)=1}$.

Now, ${{\text{e}}^{\text{y}}}\text{(x+1)=1}\Rightarrow {{\text{e}}^{\text{y}}}\text{=}\frac{\text{1}}{\text{x+1}}$.

So, taking logarithm bth sides of the equation gives

$\text{y=log}\frac{\text{1}}{\text{(x+1)}}$

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

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

That is,

$\frac{\text{dy}}{\text{dx}}=\frac{\text{-1}}{\text{x+1}}$                                                     …… (1)

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{=}\left( \frac{\text{1}}{\text{x+1}} \right)\text{=-}\left( \frac{\text{-1}}{{{\text{(x+1)}}^{\text{2}}}} \right)\text{=}\frac{\text{1}}{{{\text{(x+1)}}^{\text{2}}}} \\ 

 & \Rightarrow \frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{=}{{\left( \frac{\text{-1}}{\text{x+1}} \right)}^{\text{2}}} \\ 

\end{align}$

$\Rightarrow \frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{=}{{\left( \frac{\text{dy}}{\text{dx}} \right)}^{\text{2}}}$, using the equation (1).

Thus, it is proved that $\frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{=}{{\left( \frac{\text{dy}}{\text{dx}} \right)}^{\text{2}}}$.

Ans:  

The given equations are $\text{y=(ta}{{\text{n}}^{\text{-1}}}\text{x}{{\text{)}}^{\text{2}}}$.

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

\[\begin{align}

  & {{\text{y}}_{\text{1}}}\text{=2ta}{{\text{n}}^{\text{-1}}}\text{x}\frac{\text{d}}{\text{dx}}\text{(ta}{{\text{n}}^{\text{-1}}}\text{x)} \\ 

 & \Rightarrow {{\text{y}}_{\text{1}}}\text{=2ta}{{\text{n}}^{\text{-1}}}\text{x }\!\!\times\!\!\text{ }\frac{\text{1}}{\text{1+}{{\text{x}}^{\text{2}}}} \\ 

 & \Rightarrow \text{(1+}{{\text{x}}^{\text{2}}}\text{)}{{\text{y}}_{\text{1}}}\text{=2ta}{{\text{n}}^{\text{-1}}}\text{x} \\ 

\end{align}\]

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

$\begin{align}

  & \text{(1+}{{\text{x}}^{\text{2}}}\text{)}{{\text{y}}_{\text{2}}}\text{+2x}{{\text{y}}_{\text{1}}}\text{=2}\left( \frac{\text{1}}{\text{1+}{{\text{x}}^{\text{2}}}} \right) \\ 

 & \Rightarrow \text{(1+}{{\text{x}}^{\text{2}}}\text{)}{{\text{y}}_{\text{2}}}\text{+2x(1+}{{\text{x}}^{\text{2}}}\text{)}{{\text{y}}_{\text{1}}}\text{=2} \\ 

\end{align}$

Thus, it has been proved that $\text{(1+}{{\text{x}}^{\text{2}}}\text{)}{{\text{y}}_{\text{2}}}\text{+2x(1+}{{\text{x}}^{\text{2}}}\text{)}{{\text{y}}_{\text{1}}}\text{=2}$.