Q: Find the second order derivative of the function $\tan ^{-1} x$

Let $y=\tan ^{-1} x$ Then, $$ \begin{aligned} \frac{d y}{d x} & =\frac{d}{d x}\left(\tan ^{-1} x\right) \\ & =\frac{1}{1+x^{2}} \end{aligned} $$ Therefore, $$ \begin{aligned} \frac{d^{2} y}{d x^{2}} & =\frac{d}{d x}\left(\frac{1}{1+x^{2}}\right)=\frac{d}{d x}\left(1+x^{2}\right)^{-1} \\ & =(-1) \cdot\left(1+x^{2}\right)^{-2} \cdot \frac{d}{d x}\left(1+x^{2}\right)=\frac{-1}{\left(1+x^{2}\right)^{2}} \times 2 x \\ & =\frac{-2 x}{\left(1+x^{2}\right)^{2}} \end{aligned} $$

Question 1

Find the second order derivative of the function $\log (\log x)$

Accepted Answer

Consider, $y=\log (\log x)$

Then,

$$

\begin{aligned}

\frac{d y}{d x} & =\frac{d}{d x}[\log (\log x)] \

& =\frac{1}{\log x} \cdot \frac{d}{d x}(\log x) \

& =\frac{1}{\log x} \cdot \frac{1}{x}=(x \log x)^{-1}

\end{aligned}

$$

Therefore,

$$

\begin{aligned}

\frac{d^{2} y}{d x^{2}} & =\frac{d}{d x}\left[(x \log x)^{-1}ight] \

& =(-1) \cdot(x \log x)^{-2} \frac{d}{d x}(x \log x) \

& =\frac{-1}{(x \log x)^{2}} \cdot\left[\log x \cdot \frac{d}{d x}(x)+x \cdot \frac{d}{d x}(\log x)ight] \

& =\frac{-1}{(x \log x)^{2}} \cdot\left[\log x \cdot 1+x \cdot \frac{1}{x}ight] \

& =\frac{-(1+\log x)}{(x \log x)^{2}}

\end{aligned}

$$

Question 2

Find the second order derivative of the function $x^{2}+3 x+2$

Accepted Answer

Consider, $y=x^{2}+3 x+2$

Then,

$$

\begin{aligned}

\frac{d y}{d x} & =\frac{d}{d x}\left(x^{2}ight)+\frac{d}{d x}(3 x)+\frac{d}{d x}(2) \

& =2 x+3+0 \

& =2 x+3

\end{aligned}

$$

Therefore,

$$

\begin{aligned}

\frac{d^{2} y}{d x^{2}} & =\frac{d}{d x}(2 x+3) \

& =\frac{d}{d x}(2 x)+\frac{d}{d x}(3) \

& =2+0 \

& =2

\end{aligned}

$$

Question 3

Find the second order derivative of the function $x^{20}$

Accepted Answer

Consider, $y=x^{20}$

Then,

$$

\begin{aligned}

\frac{d y}{d x} & =\frac{d}{d x}\left(x^{20}ight) \

& =20 x^{19}

\end{aligned}

$$

Therefore,

$$

\begin{aligned}

\frac{d^{2} y}{d x^{2}} & =\frac{d}{d x}\left(20 x^{19}ight) \

& =20 \frac{d}{d x}\left(x^{19}ight) \

& =20.19 . x^{18} \

& =380 x^{18}

\end{aligned}

$$

Question 4

Find the second order derivative of the function $x \cos x$

Accepted Answer

Consider, $y=x \cos x$

Then,

$$

\begin{aligned}

\frac{d y}{d x} & =\frac{d}{d x}(x \cdot \cos x) \

& =\cos x \cdot \frac{d}{d x}(x)+x \frac{d}{d x}(\cos x) \

& =\cos x \cdot 1+x(-\sin x) \

& =\cos x-x \sin x

\end{aligned}

$$

Therefore,

$$

\begin{aligned}

\frac{d^{2} y}{d x^{2}} & =\frac{d}{d x}[\cos x-x \sin x] \

& =\frac{d}{d x}(\cos x)-\frac{d}{d x}(x \sin x) \

& =-\sin x-\left[\sin x \cdot \frac{d}{d x}(x)+x \cdot \frac{d}{d x}(\sin x)ight] \

& =-\sin x-(\sin x+x \cos x) \

& =-(x \cos x+2 \sin x)

\end{aligned}

$$

Question 5

Find the second order derivative of the function $\log x$

Accepted Answer

Let $y=\log x$

Then,

$$

\begin{aligned}

\frac{d y}{d x} & =\frac{d}{d x}(\log x) \

& =\frac{1}{x}

\end{aligned}

$$

Therefore,

$$

\begin{aligned}

\frac{d^{2} y}{d x^{2}} & =\frac{d}{d x}\left(\frac{1}{x}ight) \

& =\frac{-1}{x^{2}}

\end{aligned}

$$

Question 6

Find the second order derivative of the function $x^{3} \log x$

Accepted Answer

Let $y=x^{3} \log x$

Then,

$$

\begin{aligned}

\frac{d y}{d x} & =\frac{d}{d x}\left[x^{3} \log xight] \

& =\log x \cdot \frac{d}{d x}\left(x^{3}ight)+x^{3} \cdot \frac{d}{d x}(\log x) \

& =\log x \cdot 3 x^{2}+x^{3} \cdot \frac{1}{x}=\log x \cdot 3 x^{2}+x^{2} \

& =x^{2}(1+3 \log x)

\end{aligned}

$$

Therefore,

$$

\begin{aligned}

\frac{d^{2} y}{d x^{2}} & =\frac{d}{d x}\left[x^{2}(1+3 \log x)ight] \

& =(1+3 \log x) \cdot \frac{d}{d x}\left(x^{2}ight)+x^{2} \frac{d}{d x}(1+3 \log x) \

& =(1+3 \log x) \cdot 2 x+x^{2} \cdot \frac{3}{x} \

& =2 x+6 \log x+3 x \

& =5 x+6 x \log x \

& =x(5+6 \log x)

\end{aligned}

$$

Question 7

Find the second order derivative of the function $e^{x} \sin 5 x$

Accepted Answer

Let $y=e^{x} \sin 5 x$

Then,

$$

\begin{aligned}

\frac{d y}{d x} & =\frac{d}{d x}\left(e^{x} \sin 5 xight) \

& =\sin 5 x 	imes \frac{d}{d x}\left(e^{x}ight)+e^{x} \frac{d}{d x}(\sin 5 x) \

& =\sin 5 x \cdot e^{x}+e^{x} \cdot \cos 5 x \cdot \frac{d}{d x}(5 x) \

& =e^{x} \sin 5 x+e^{x} \cos 5 x \cdot 5 \

& =e^{x}(\sin 5 x+5 \cos 5 x)

\end{aligned}

$$

Therefore,

$$

\begin{aligned}

\frac{d^{2} y}{d x^{2}} & =\frac{d}{d x}\left[e^{x}(\sin 5 x+5 \cos 5 x)ight] \

& =(\sin 5 x+5 \cos 5 x) \cdot \frac{d}{d x}\left(e^{x}ight)+e^{x} \cdot \frac{d}{d x}(\sin 5 x+5 \cos 5 x) \

& =(\sin 5 x+5 \cos 5 x) e^{x}+e^{x}\left[\cos 5 x \cdot \frac{d}{d x}(5 x)+5(-\sin 5 x) \cdot \frac{d}{d x}(5 x)ight] \

& =e^{x}(\sin 5 x+5 \cos 5 x)+e^{x}(5 \cos 5 x-25 \sin 5 x) \

& =e^{x}(10 \cos 5 x-24 \sin 5 x) \

& =2 e^{x}(5 \cos 5 x-12 \sin 5 x)

\end{aligned}

$$

Question 8

Find the second order derivative of the function $e^{6 x} \cos 3 x$

Accepted Answer

Let $y=e^{6 x} \cos 3 x$

Then,

$$

\begin{align*}

\frac{d y}{d x} & =\frac{d}{d x}\left(e^{6 x} \cos 3 xight)=\cos 3 x \cdot \frac{d}{d x}\left(e^{6 x}ight)+e^{6 x} \cdot \frac{d}{d x}(\cos 3 x) \

& =\cos 3 x \cdot e^{6 x} \cdot \frac{d}{d x}(6 x)+e^{6 x} \cdot(-\sin 3 x) \cdot \frac{d}{d x}(3 x) \

& =6 e^{6 x} \cos 3 x-3 e^{6 x} \sin 3 x 	ag{1}

\end{align*}

$$

Therefore,

$$

\begin{aligned}

\frac{d^{2} y}{d x^{2}} & =\frac{d}{d x}\left(6 e^{6 x} \cos 3 x-3 e^{6 x} \sin 3 xight)=6 \cdot \frac{d}{d x}\left(e^{6 x} \cos 3 xight)-3 \cdot \frac{d}{d x}\left(e^{6 x} \sin 3 xight) \

& =6 \cdot\left[6 e^{6 x} \cos 3 x-3 e^{6 x} \sin 3 xight]-3 \cdot\left[\sin 3 x \cdot \frac{d}{d x}\left(e^{6 x}ight)+e^{6 x} \cdot \frac{d}{d x}(\sin 3 x)ight] \quad[	ext { using }(1)] \

& =36 e^{6 x} \cos 3 x-18 e^{6 x} \sin 3 x-3\left[\sin 3 x \cdot e^{6 x} \cdot 6+e^{6 x} \cdot \cos 3 x \cdot 3ight] \

& =36 e^{6 x} \cos 3 x-18 e^{6 x} \sin 3 x-18 e^{6 x} \sin 3 x-9 e^{6 x} \cos 3 x \

& =27 e^{6 x} \cos 3 x-36 e^{6 x} \sin 3 x \

& =9 e^{6 x}(3 \cos 3 x-4 \sin 3 x)

\end{aligned}

$$

Question 9

Find the second order derivative of the function $	an ^{-1} x$

Accepted Answer

Let $y=	an ^{-1} x$

Then,

$$

\begin{aligned}

\frac{d y}{d x} & =\frac{d}{d x}\left(	an ^{-1} xight) \

& =\frac{1}{1+x^{2}}

\end{aligned}

$$

Therefore,

$$

\begin{aligned}

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

& =(-1) \cdot\left(1+x^{2}ight)^{-2} \cdot \frac{d}{d x}\left(1+x^{2}ight)=\frac{-1}{\left(1+x^{2}ight)^{2}} 	imes 2 x \

& =\frac{-2 x}{\left(1+x^{2}ight)^{2}}

\end{aligned}

$$

Question 10

Find the second order derivative of the function $\sin (\log x)$

Accepted Answer

Let $y=\sin (\log x)$

Then,

$$

\begin{aligned}

\frac{d y}{d x} & =\frac{d}{d x}[\sin x(\log x)] \

& =\cos (\log x) \cdot \frac{d}{d x}(\log x) \

& =\frac{\cos (\log x)}{x}

\end{aligned}

$$

Therefore,

$$

\begin{aligned}

\frac{d^{2} y}{d x^{2}} & =\frac{d}{d x}\left[\frac{\cos (\log x)}{x}ight] \

& =\frac{x \cdot \frac{d}{d x}[\cos (\log x)]-\cos (\log x) \cdot \frac{d}{d x}(x)}{x^{2}}

\end{aligned}

$$

\begin{aligned}

\frac{d^{2} y}{d x^{2}} & =\frac{x\left[-\sin (\log x) \cdot \frac{d}{d x}(\log x)ight]-\cos (\log x) \cdot 1}{x^{2}} \

& =\frac{-x \sin (\log x) \cdot \frac{1}{x}-\cos (\log x)}{x^{2}} \

& =\frac{-[\sin (\log x)+\cos (\log x)]}{x^{2}}

\end{aligned}

$$

Question 11

If $y=5 \cos x-3 \sin x$, prove that $\frac{d^{2} y}{d x^{2}}+y=0$

Accepted Answer

Given, $y=5 \cos x-3 \sin x$

Then,

$$

\begin{aligned}

\frac{d y}{d x} & =\frac{d}{d x}(5 \cos x)-\frac{d}{d x}(3 \sin x) \

& =5 \frac{d}{d x}(\cos x)-3 \frac{d}{d x}(\sin x) \

& =5(-\sin x)-3 \cos x \

& =-(5 \sin x+3 \cos x)

\end{aligned}

$$

Therefore,

$$

\begin{aligned}

\frac{d^{2} y}{d x^{2}} & =\frac{d}{d x}[-(5 \sin x+3 \cos x)] \

& =-\left[5 \cdot \frac{d}{d x}(\sin x)+3 \cdot \frac{d}{d x}(\cos x)ight] \

& =-[5 \cos x+3(-\sin x)] \

& =-[5 \cos x-3 \sin x] \

& =-y

\end{aligned}

$$

Thus, $\frac{d^{2} y}{d x^{2}}+y=0$

Hence proved.

Question 12

If $y=\cos ^{-1} x$, find $\frac{d^{2} y}{d x^{2}}$ in terms of $y$ alone.

Accepted Answer

Given, $y=\cos ^{-1} x$

Then,

$$

\begin{aligned}

\frac{d y}{d x} & =\frac{d}{d x}\left(\cos ^{-1} xight) \

& =\frac{-1}{\sqrt{1-x^{2}}} \

& =-\left(1-x^{2}ight)^{\frac{-1}{2}}

\end{aligned}

$$

Therefore,

$$

\begin{align*}

\frac{d^{2} y}{d x^{2}} & =\frac{d}{d x}\left[-\left(1-x^{2}ight)^{\frac{-1}{2}}ight] \

& =-\left(-\frac{1}{2}ight) \cdot\left(1-x^{2}ight)^{\frac{-3}{2}} \cdot \frac{d}{d x}\left(1-x^{2}ight) \

& =\frac{1}{2 \sqrt{\left(1-x^{2}ight)^{3}}} 	imes(-2 x) \

\frac{d^{2} y}{d x^{2}} & =\frac{-x}{\sqrt{\left(1-x^{2}ight)^{3}}} 	ag{1}

\end{align*}

$$

But we need to calculate $\frac{d^{2} y}{d x^{2}}$ in terms of $y$

$$

\begin{aligned}

& \Rightarrow y=\cos ^{-1} x \

& \Rightarrow x=\cos y

\end{aligned}

$$

Putting $x=\cos y$ in equation (1), we get

$$

\begin{aligned}

\frac{d^{2} y}{d x^{2}} & =\frac{-\cos y}{\sqrt{\left(1-\cos ^{2} yight)^{3}}} \

& =\frac{-\cos y}{\sqrt{\left(\sin ^{2} yight)^{3}}} \

& =\frac{-\cos y}{\sin ^{3} y} \

& =\frac{-\cos y}{\sin y} 	imes \frac{1}{\sin ^{2} y} \

& =-\cot y \cdot \operatorname{cosec}^{2} y

\end{aligned}

$$

Question 13

If $y=3 \cos (\log x)+4 \sin (\log x)$, show that $x^{2} y_{2}+x y_{1}+y=0$

Accepted Answer

Given, $y=3 \cos (\log x)+4 \sin (\log x)$

Then,

$$

\begin{aligned}

y_{1} & =3 \cdot \frac{d}{d x}[\cos (\log x)]+4 \cdot \frac{d}{d x}[\sin (\log x)] \

& =3 \cdot\left[-\sin (\log x) \cdot \frac{d}{d x}(\log x)ight]+4 \cdot\left[\cos (\log x) \cdot \frac{d}{d x}(\log x)ight] \

& =\frac{-3 \sin (\log x)}{x}+\frac{4 \cos (\log x)}{x} \

& =\frac{4 \cos (\log x)-3 \sin (\log x)}{x}

\end{aligned}

$$

Therefore,

$$

\begin{aligned}

y_{2} & =\frac{d}{d x}\left(\frac{4 \cos (\log x)-3 \sin (\log x)}{x}ight) \

& =\frac{x \cdot\{4 \cos (\log x)-3 \sin (\log x)\}^{\prime}-\{4 \cos (\log x)-3 \sin (\log x)\}\{x\}^{\prime}}{x^{2}} \

& =\frac{x \cdot\left[4\{\cos (\log x)\}^{\prime}-\{3 \sin (\log x)\}^{\prime}ight]-\{4 \cos (\log x)-3 \sin (\log x)\} \cdot 1}{x^{2}} \

& =\frac{x \cdot\left[-4 \sin (\log x) \cdot(\log x)^{\prime}-3 \cos (\log x) \cdot(\log x)^{\prime}ight]-4 \cos (\log x)+3 \sin (\log x)}{x^{2}} \

& =\frac{x \cdot\left[-4 \sin (\log x) \frac{1}{x}-3 \cos (\log x) \frac{1}{x}ight]-4 \cos (\log x)+3 \sin (\log x)}{x^{2}} \

& =\frac{-4 \sin (\log x)-3 \cos (\log x)-4 \cos (\log x)+3 \sin (\log x)}{x^{2}} \

& =\frac{-\sin (\log x)-7 \cos (\log x)}{x^{2}}

\end{aligned}

$$

Thus,

$$

\begin{aligned}

x^{2} y_{2}+x y_{1}+y & =\left[\begin{array}{c}

x^{2}\left(\frac{-\sin (\log x)-7 \cos (\log x)}{x^{2}}ight)+x\left(\frac{4 \cos (\log x)-3 \sin (\log x)}{x}ight) \

+3 \cos (\log x)+4 \sin (\log x)

\end{array}ight] \

& =\left[\begin{array}{c}

-\sin (\log x)-7 \cos (\log x)+4 \cos (\log x)-3 \sin (\log x) \

+3 \cos (\log x)+4 \sin (\log x)

\end{array}ight] \

& =0

\end{aligned}

$$

Hence proved.

Question 14

If $y=A e^{m x}+B e^{n x}$, show that $\frac{d^{2} y}{d x^{2}}-(m+n) \frac{d y}{d x}+m n y=0$.

Accepted Answer

Given, $y=A e^{m x}+B e^{n x}$

Then,

$$

\begin{aligned}

\frac{d y}{d x} & =A \cdot \frac{d}{d x}\left(e^{m x}ight)+B \cdot \frac{d}{d x}\left(e^{n x}ight) \

& =A \cdot e^{m x} \cdot \frac{d}{d x}(m x)+B \cdot e^{n x} \cdot \frac{d}{d x}(n x) \

& =A m e^{m x}+B n e^{n x}

\end{aligned}

$$

Therefore,

$$

\begin{aligned}

\frac{d^{2} y}{d x^{2}} & =\frac{d}{d x}\left(A m e^{m x}+B n e^{n x}ight) \

& =A m \cdot \frac{d}{d x}\left(e^{m x}ight)+B n \cdot \frac{d}{d x}\left(e^{n x}ight) \

& =A m \cdot e^{m x} \cdot \frac{d}{d x}(m x)+B n \cdot e^{n x} \cdot \frac{d}{d x}(n x) \

& =A m^{2} e^{m x}+B n^{2} e^{n x}

\end{aligned}

$$

Thus,

$$

\begin{aligned}

\frac{d^{2} y}{d x^{2}}-(m+n) \frac{d y}{d x}+m n y & =A m^{2} e^{m x}+B n^{2} e^{n x}-(m+n) \cdot\left(A m e^{m x}+B n e^{n x}ight)+m n\left(A e^{m x}+B e^{n x}ight) \

& =A m^{2} e^{m x}+B n^{2} e^{n x}-A m^{2} e^{m x}-B m n e^{n x}-A m n e^{m x}-B n^{2} e^{n x}+A m n e^{m x}+B m n e^{n x} \

& =0

\end{aligned}

$$

Hence proved.

Question 15

If $y=500 e^{7 x}+600 e^{-7 x}$, show that $\frac{d^{2} y}{d x^{2}}=49 y$

Accepted Answer

Given, $y=500 e^{7 x}+600 e^{-7 x}$

Then,

$$

\begin{aligned}

\frac{d y}{d x} & =500 \cdot \frac{d}{d x}\left(e^{7 x}ight)+600 \cdot \frac{d}{d x}\left(e^{-7 x}ight) \

& =500 \cdot e^{7 x} \cdot \frac{d}{d x}(7 x)+600 \cdot e^{-7 x} \cdot \frac{d}{d x}(-7 x) \

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

\end{aligned}

$$

Therefore,

$$

\begin{aligned}

\frac{d^{2} y}{d x^{2}} & =3500 e^{7 x} \cdot \frac{d}{d x}\left(e^{7 x}ight)-4200 \cdot \frac{d}{d x}\left(e^{-7 x}ight) \

& =3500 \cdot e^{7 x} \cdot \frac{d}{d x}(7 x)-4200 \cdot e^{-7 x} \cdot \frac{d}{d x}(-7 x) \

& =7 	imes 3500 \cdot e^{7 x}+7 	imes 4200 \cdot e^{-7 x} \

& =49 	imes 500 \cdot e^{7 x}+49 	imes 600 e^{-7 x} \

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

& =49 y

\end{aligned}

$$

Hence proved.

Question 16

If $e^{y}(x+1)=1$, show that $\frac{d^{2} y}{d x^{2}}=\left(\frac{d y}{d x}\right)^{2}$

Accepted Answer

Given, $e^{y}(x+1)=1$

$\Rightarrow e^{y}(x+1)=1$

$\Rightarrow e^{y}=\frac{1}{x+1}$

Taking log on both sides, we get

$$

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

$$

Differentiating with respect to $x$, we get

$$

\begin{aligned}

\frac{d y}{d x} & =(x+1) \frac{d}{d x}\left(\frac{1}{x+1}ight) \

& =(x+1) \cdot \frac{-1}{(x+1)^{2}} \

& =\frac{-1}{x+1}

\end{aligned}

$$

Therefore,

$$

\begin{aligned}

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

& =\frac{1}{(x+1)^{2}}=\left(\frac{-1}{x+1}ight)^{2} \

& =\left(\frac{d y}{d x}ight)^{2}

\end{aligned}

$$

Hence proved.

Question 17

If $y=\left(	an ^{-1} xight)^{2}$, show that $\left(x^{2}+1ight)^{2} y_{2}+2 x\left(x^{2}+1ight) y_{1}=2$

Accepted Answer

Given, $y=\left(	an ^{-1} xight)^{2}$

Then,

$$

\begin{aligned}

& \Rightarrow y_{1}=2 	an ^{-1} x \frac{d}{d x}\left(	an ^{-1} xight) \

& \Rightarrow y_{1}=2 	an ^{-1} x \cdot\left(\frac{1}{1+x^{2}}ight) \

& \Rightarrow\left(1+x^{2}ight) y_{1}=2 	an ^{-1} x

\end{aligned}

$$

Again, differentiating with respect to $x$, we get

$$

\begin{aligned}

& \Rightarrow\left(1+x^{2}ight) y_{2}+2 x y_{1}=2\left(\frac{1}{1+x^{2}}ight) \

& \Rightarrow\left(1+x^{2}ight)^{2} y_{2}+2 x\left(1+x^{2}ight) y_{1}=2

\end{aligned}

$$

Hence proved.

\section*{EXERCISE 5.8}