# NCERT Solutions For Class 12 Maths Chapter 5 Exercise 5.7

Go back to  'Continuity and Differentiability'

## Chapter 5 Ex.5.7 Question 1

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

### Solution

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

Then,

\begin{align}\frac{{dy}}{{dx}} &= \frac{d}{{dx}}\left( {{x^2}} \right) + \frac{d}{{dx}}\left( {3x} \right) + \frac{d}{{dx}}\left( 2 \right)\\ &= 2x + 3 + 0\\ &= 2x + 3\end{align}

Therefore,

\begin{align}\frac{{{d^2}y}}{{d{x^2}}} &= \frac{d}{{dx}}\left( {2x + 3} \right)\\ &= \frac{d}{{dx}}\left( {2x} \right) + \frac{d}{{dx}}\left( 3 \right)\\& = 2 + 0\\& = 2\end{align}

## Chapter 5 Ex.5.7 Question 2

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

### Solution

Consider, $$y = {x^{20}}$$

Then,

\begin{align}\frac{{dy}}{{dx}} &= \frac{d}{{dx}}\left( {{x^{20}}} \right)\\ &= 20{x^{19}}\end{align}

Therefore,

\begin{align}\frac{{{d^2}y}}{{d{x^2}}}&= \frac{d}{{dx}}\left( {20{x^{19}}} \right)\\ &= 20\frac{d}{{dx}}\left( {{x^{19}}} \right)\\ &= 20.19.{x^{18}}\\& = 380{x^{18}}\end{align}

## Chapter 5 Ex.5.7 Question 3

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

### Solution

Consider, $$y = x\cos x$$

Then,

\begin{align}\frac{{dy}}{{dx}} &= \frac{d}{{dx}}\left( {x.\cos x} \right)\\ &= \cos x.\frac{d}{{dx}}\left( x \right) + x\frac{d}{{dx}}\left( {\cos x} \right)\\& = \cos x.1 + x\left( { - \sin x} \right)\\& = \cos x - x\sin x\end{align}

Therefore,

\begin{align}\frac{{{d^2}y}}{{d{x^2}}}& = \frac{d}{{dx}}\left[ {\cos x - x\sin x} \right]\\ &= \frac{d}{{dx}}\left( {\cos x} \right) - \frac{d}{{dx}}\left( {x\sin x} \right)\\ &= - \sin x - \left[ {\sin x.\frac{d}{{dx}}\left( x \right) + x.\frac{d}{{dx}}\left( {\sin x} \right)} \right]\\ &= - \sin x - \left( {\sin x + x\cos x} \right)\\& = - \left( {x\cos x + 2\sin x} \right)\end{align}

## Chapter 5 Ex.5.7 Question 4

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

### Solution

Let $$y = \log x$$

Then,

\begin{align}\frac{{dy}}{{dx}} &= \frac{d}{{dx}}\left( {\log x} \right)\\ &= \frac{1}{x}\end{align}

Therefore,

\begin{align}\frac{{{d^2}y}}{{d{x^2}}} &= \frac{d}{{dx}}\left( {\frac{1}{x}} \right)\\& = \frac{{ - 1}}{{{x^2}}}\end{align}

## Chapter 5 Ex.5.7 Question 5

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

### Solution

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

Then,

\begin{align}\frac{{dy}}{{dx}}& = \frac{d}{{dx}}\left[ {{x^3}\log x} \right]\\& = \log x.\frac{d}{{dx}}\left( {{x^3}} \right) + {x^3}.\frac{d}{{dx}}\left( {\log x} \right)\\& = \log x.3{x^2} + {x^3}.\frac{1}{x} = \log x.3{x^2} + {x^2}\\& = {x^2}\left( {1 + 3\log x} \right)\end{align}

Therefore,

\begin{align}\frac{{{d^2}y}}{{d{x^2}}} &= \frac{d}{{dx}}\left[ {{x^2}\left( {1 + 3\log x} \right)} \right]\\ &= \left( {1 + 3\log x} \right).\frac{d}{{dx}}\left( {{x^2}} \right) + {x^2}\frac{d}{{dx}}\left( {1 + 3\log x} \right)\\& = \left( {1 + 3\log x} \right).2x + {x^2}.\frac{3}{x}\\& = 2x + 6\log x + 3x\\ &= 5x + 6x\log x\\ &= x\left( {5 + 6\log x} \right)\end{align}

## Chapter 5 Ex.5.7 Question 6

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

### Solution

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

Then,

\begin{align}\frac{{dy}}{{dx}} &= \frac{d}{{dx}}\left( {{e^x}\sin 5x} \right)\\ &= \sin 5x \times \frac{d}{{dx}}\left( {{e^x}} \right) + {e^x}\frac{d}{{dx}}\left( {\sin 5x} \right)\\ &= \sin 5x.{e^x} + {e^x}.\cos 5x.\frac{d}{{dx}}\left( {5x} \right)\\ &= {e^x}\sin 5x + {e^x}\cos 5x.5\\ &= {e^x}\left( {\sin 5x + 5\cos 5x} \right)\end{align}

Therefore,

\begin{align}\frac{{{d^2}y}}{{d{x^2}}} &= \frac{d}{{dx}}\left[ {{e^x}\left( {\sin 5x + 5\cos 5x} \right)} \right]\\ &= \left( {\sin 5x + 5\cos 5x} \right).\frac{d}{{dx}}\left( {{e^x}} \right) + {e^x}.\frac{d}{{dx}}\left( {\sin 5x + 5\cos 5x} \right)\\& = \left( {\sin 5x + 5\cos 5x} \right){e^x} + {e^x}\left[ {\cos 5x.\frac{d}{{dx}}\left( {5x} \right) + 5\left( { - \sin 5x} \right).\frac{d}{{dx}}\left( {5x} \right)} \right]\\ &= {e^x}\left( {\sin 5x + 5\cos 5x} \right) + {e^x}\left( {5\cos 5x - 25\sin 5x} \right)\\& = {e^x}\left( {10\cos 5x - 24\sin 5x} \right)\\ &= 2{e^x}\left( {5\cos 5x - 12\sin 5x} \right)\end{align}

## Chapter 5 Ex.5.7 Question 7

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

### Solution

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

Then,

\begin{align}\frac{{dy}}{{dx}} &= \frac{d}{{dx}}\left( {{e^{6x}}\cos 3x} \right) = \cos 3x.\frac{d}{{dx}}\left( {{e^{6x}}} \right) + {e^{6x}}.\frac{d}{{dx}}\left( {\cos 3x} \right)\\ &= \cos 3x.{e^{6x}}.\frac{d}{{dx}}\left( {6x} \right) + {e^{6x}}.\left( { - \sin 3x} \right).\frac{d}{{dx}}\left( {3x} \right)\\ &= 6{e^{6x}}\cos 3x - 3{e^{6x}}\sin 3x \qquad \qquad \ldots \left( 1 \right)\end{align}

Therefore,

\begin{align}\frac{{{d^2}y}}{{d{x^2}}}& = \frac{d}{{dx}}\left( {6{e^{6x}}\cos 3x - 3{e^{6x}}\sin 3x} \right) = 6.\frac{d}{{dx}}\left( {{e^{6x}}\cos 3x} \right) - 3.\frac{d}{{dx}}\left( {{e^{6x}}\sin 3x} \right)\\& = 6.\left[ {6{e^{6x}}\cos 3x - 3{e^{6x}}\sin 3x} \right] - 3.\left[ {\sin 3x.\frac{d}{{dx}}\left( {{e^{6x}}} \right) + {e^{6x}}.\frac{d}{{dx}}\left( {\sin 3x} \right)} \right] \quad \left[ {{\text{using }}\left( 1 \right)} \right]\\ &= 36{e^{6x}}\cos 3x - 18{e^{6x}}\sin 3x - 3\left[ {\sin 3x.{e^{6x}}.6 + {e^{6x}}.\cos 3x.3} \right]\\ &= 36{e^{6x}}\cos 3x - 18{e^{6x}}\sin 3x - 18{e^{6x}}\sin 3x - 9{e^{6x}}\cos 3x\\ &= 27{e^{6x}}\cos 3x - 36{e^{6x}}\sin 3x\\& = 9{e^{6x}}\left( {3\cos 3x - 4\sin 3x} \right)\end{align}

## Chapter 5 Ex.5.7 Question 8

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

### Solution

Let $$y = {\tan ^{ - 1}}x$$

Then,

\begin{align}\frac{{dy}}{{dx}} &= \frac{d}{{dx}}\left( {{{\tan }^{ - 1}}x} \right)\\& = \frac{1}{{1 + {x^2}}}\end{align}

Therefore,

\begin{align}\frac{{{d^2}y}}{{d{x^2}}} &= \frac{d}{{dx}}\left( {\frac{1}{{1 + {x^2}}}} \right) = \frac{d}{{dx}}{\left( {1 + {x^2}} \right)^{ - 1}}\\& = \left( { - 1} \right).{\left( {1 + {x^2}} \right)^{ - 2}}.\frac{d}{{dx}}\left( {1 + {x^2}} \right) = \frac{{ - 1}}{{{{\left( {1 + {x^2}} \right)}^2}}} \times 2x\\ &= \frac{{ - 2x}}{{{{\left( {1 + {x^2}} \right)}^2}}}\end{align}

## Chapter 5 Ex.5.7 Question 9

Find the second order derivative of the function $$\log \left( {\log x} \right)$$

### Solution

Consider, $$y = \log \left( {\log x} \right)$$

Then,

\begin{align}\frac{{dy}}{{dx}} &= \frac{d}{{dx}}\left[ {\log \left( {\log x} \right)} \right]\\ &= \frac{1}{{\log x}}.\frac{d}{{dx}}\left( {\log x} \right)\\& = \frac{1}{{\log x}}.\frac{1}{x} = {\left( {x\log x} \right)^{ - 1}}\end{align}

Therefore,

\begin{align}\frac{{{d^2}y}}{{d{x^2}}} &= \frac{d}{{dx}}\left[ {{{\left( {x\log x} \right)}^{ - 1}}} \right]\\& = \left( { - 1} \right).{\left( {x\log x} \right)^{ - 2}}\frac{d}{{dx}}\left( {x\log x} \right)\\ &= \frac{{ - 1}}{{{{\left( {x\log x} \right)}^2}}}.\left[ {\log x.\frac{d}{{dx}}\left( x \right) + x.\frac{d}{{dx}}\left( {\log x} \right)} \right]\\ &= \frac{{ - 1}}{{{{\left( {x\log x} \right)}^2}}}.\left[ {\log x.1 + x.\frac{1}{x}} \right]\\ &= \frac{{ - \left( {1 + \log x} \right)}}{{{{\left( {x\log x} \right)}^2}}}\end{align}

## Chapter 5 Ex.5.7 Question 10

Find the second order derivative of the function $$\sin \left( {\log x} \right)$$

### Solution

Let $$y = \sin \left( {\log x} \right)$$

Then,

\begin{align}\frac{{dy}}{{dx}} &= \frac{d}{{dx}}\left[ {\sin x\left( {\log x} \right)} \right]\\ &= \cos \left( {\log x} \right).\frac{d}{{dx}}\left( {\log x} \right)\\& = \frac{{\cos \left( {\log x} \right)}}{x}\end{align}

Therefore,

\begin{align}\frac{{{d^2}y}}{{d{x^2}}}&= \frac{d}{{dx}}\left[ {\frac{{\cos \left( {\log x} \right)}}{x}} \right]\\ &= \frac{{x.\frac{d}{{dx}}\left[ {\cos \left( {\log x} \right)} \right] - \cos \left( {\log x} \right).\frac{d}{{dx}}\left( x \right)}}{{{x^2}}}\end{align}

\begin{align}\frac{{{d^2}y}}{{d{x^2}}} &= \frac{{x\left[ { - \sin \left( {\log x} \right).\frac{d}{{dx}}\left( {\log x} \right)} \right] - \cos \left( {\log x} \right).1}}{{{x^2}}}\\ &= \frac{{ - x\sin \left( {\log x} \right).\frac{1}{x} - \cos \left( {\log x} \right)}}{{{x^2}}}\\ &= \frac{{ - \left[ {\sin \left( {\log x} \right) + \cos \left( {\log x} \right)} \right]}}{{{x^2}}}\end{align}

## Chapter 5 Ex.5.7 Question 11

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

### Solution

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

Then,

\begin{align}\frac{{dy}}{{dx}}& = \frac{d}{{dx}}\left( {5\cos x} \right) - \frac{d}{{dx}}\left( {3\sin x} \right)\\ &= 5\frac{d}{{dx}}\left( {\cos x} \right) - 3\frac{d}{{dx}}\left( {\sin x} \right)\\ &= 5\left( { - \sin x} \right) - 3\cos x\\ &= - \left( {5\sin x + 3\cos x} \right)\end{align}

Therefore,

\begin{align}\frac{{{d^2}y}}{{d{x^2}}} &= \frac{d}{{dx}}\left[ { - \left( {5\sin x + 3\cos x} \right)} \right]\\& = - \left[ {5.\frac{d}{{dx}}\left( {\sin x} \right) + 3.\frac{d}{{dx}}\left( {\cos x} \right)} \right]\\ &= - \left[ {5\cos x + 3\left( { - \sin x} \right)} \right]\\ &= - \left[ {5\cos x - 3\sin x} \right]\\ &= - y\end{align}

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

Hence proved.

## Chapter 5 Ex.5.7 Question 12

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

### Solution

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

Then,

\begin{align}\frac{{dy}}{{dx}}& = \frac{d}{{dx}}\left( {{{\cos }^{ - 1}}x} \right)\\ &= \frac{{ - 1}}{{\sqrt {1 - {x^2}} }}\\ &= - {\left( {1 - {x^2}} \right)^{\frac{{ - 1}}{2}}}\end{align}

Therefore,

\begin{align}\frac{{{d^2}y}}{{d{x^2}}} &= \frac{d}{{dx}}\left[ { - {{\left( {1 - {x^2}} \right)}^{\frac{{ - 1}}{2}}}} \right]\\ &= - \left( { - \frac{1}{2}} \right).{\left( {1 - {x^2}} \right)^{\frac{{ - 3}}{2}}}.\frac{d}{{dx}}\left( {1 - {x^2}} \right)\\ &= \frac{1}{{2\sqrt {{{\left( {1 - {x^2}} \right)}^3}} }} \times \left( { - 2x} \right)\\\frac{{{d^2}y}}{{d{x^2}}} &= \frac{{ - x}}{{\sqrt {{{\left( {1 - {x^2}} \right)}^3}} }} \qquad \ldots \left( 1 \right)\end{align}

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

\begin{align} &\Rightarrow \; y = {\cos ^{ - 1}}x\\& \Rightarrow \; x = \cos y\end{align}

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

\begin{align}\frac{{{d^2}y}}{{d{x^2}}} &= \frac{{ - \cos y}}{{\sqrt {{{\left( {1 - {{\cos }^2}y} \right)}^3}} }}\\& = \frac{{ - \cos y}}{{\sqrt {{{\left( {{{\sin }^2}y} \right)}^3}} }}\\ &= \frac{{ - \cos y}}{{{{\sin }^3}y}}\\& = \frac{{ - \cos y}}{{\sin y}} \times \frac{1}{{{{\sin }^2}y}}\\& = - \cot y.{{\mathop{\text cosec}\nolimits} ^2}y\end{align}

## Chapter 5 Ex.5.7 Question 13

If $$y = 3\cos \left( {\log x} \right) + 4\sin \left( {\log x} \right)$$, show that $${x^2}{y_2} + x{y_1} + y = 0$$

### Solution

Given, $$y = 3\cos \left( {\log x} \right) + 4\sin \left( {\log x} \right)$$

Then,

\begin{align}{y_1} &= 3.\frac{d}{{dx}}\left[ {\cos \left( {\log x} \right)} \right] + 4.\frac{d}{{dx}}\left[ {\sin \left( {\log x} \right)} \right]\\ &= 3.\left[ { - \sin \left( {\log x} \right).\frac{d}{{dx}}\left( {\log x} \right)} \right] + 4.\left[ {\cos \left( {\log x} \right).\frac{d}{{dx}}\left( {\log x} \right)} \right]\\ &= \frac{{ - 3\sin \left( {\log x} \right)}}{x} + \frac{{4\cos \left( {\log x} \right)}}{x}\\ &= \frac{{4\cos \left( {\log x} \right) - 3\sin \left( {\log x} \right)}}{x}\end{align}

Therefore,

\begin{align}{y_2} &= \frac{d}{{dx}}\left( {\frac{{4\cos \left( {\log x} \right) - 3\sin \left( {\log x} \right)}}{x}} \right)\\ &= \frac{{x.\left\{ {4\cos \left( {\log x} \right) - 3\sin \left( {\log x} \right)} \right\}' - \left\{ {4\cos \left( {\log x} \right) - 3\sin \left( {\log x} \right)} \right\}\left\{ x \right\}'}}{{{x^2}}}\\ &= \frac{{x.\left[ {4\left\{ {\cos \left( {\log x} \right)} \right\}' - \left\{ {3\sin \left( {\log x} \right)} \right\}'} \right] - \left\{ {4\cos \left( {\log x} \right) - 3\sin \left( {\log x} \right)} \right\}.1}}{{{x^2}}}\\ &= \frac{{x.\left[ { - 4\sin \left( {\log x} \right).\left( {\log x} \right)' - 3\cos \left( {\log x} \right).\left( {\log x} \right)'} \right] - 4\cos \left( {\log x} \right) + 3\sin \left( {\log x} \right)}}{{{x^2}}}\\& = \frac{{x.\left[ { - 4\sin \left( {\log x} \right)\frac{1}{x} - 3\cos \left( {\log x} \right)\frac{1}{x}} \right] - 4\cos \left( {\log x} \right) + 3\sin \left( {\log x} \right)}}{{{x^2}}}\\& = \frac{{ - 4\sin \left( {\log x} \right) - 3\cos \left( {\log x} \right) - 4\cos \left( {\log x} \right) + 3\sin \left( {\log x} \right)}}{{{x^2}}}\\ &= \frac{{ - \sin \left( {\log x} \right) - 7\cos \left( {\log x} \right)}}{{{x^2}}}\end{align}

Thus,

\begin{align}{x^2}{y_2} + x{y_1} + y &= \left[ \begin{array}{l}{x^2}\left( {\frac{{ - \sin \left( {\log x} \right) - 7\cos \left( {\log x} \right)}}{{{x^2}}}} \right) + x\left( {\frac{{4\cos \left( {\log x} \right) - 3\sin \left( {\log x} \right)}}{x}} \right)\\ + 3\cos \left( {\log x} \right) + 4\sin \left( {\log x} \right)\end{array} \right]\\[10pt]& = \left[ \begin{array}{l} - \sin \left( {\log x} \right) - 7\cos \left( {\log x} \right) + 4\cos \left( {\log x} \right) - 3\sin \left( {\log x} \right)\\+ 3\cos \left( {\log x} \right) + 4\sin \left( {\log x} \right)\end{array} \right]\\[10pt]&= 0\end{align}

Hence proved.

## Chapter 5 Ex.5.7 Question 14

If $$y = A{e^{mx}} + B{e^{nx}}$$, show that $$\frac{{{d^2}y}}{{d{x^2}}} - \left( {m + n} \right)\frac{{dy}}{{dx}} + mny = 0$$.

### Solution

Given, $$y = A{e^{mx}} + B{e^{nx}}$$

Then,

\begin{align}\frac{{dy}}{{dx}} &= A.\frac{d}{{dx}}\left( {{e^{mx}}} \right) + B.\frac{d}{{dx}}\left( {{e^{nx}}} \right)\\& = A.{e^{mx}}.\frac{d}{{dx}}\left( {mx} \right) + B.{e^{nx}}.\frac{d}{{dx}}\left( {nx} \right)\\& = Am{e^{mx}} + Bn{e^{nx}}\end{align}

Therefore,

\begin{align}\frac{{{d^2}y}}{{d{x^2}}} &= \frac{d}{{dx}}\left( {Am{e^{mx}} + Bn{e^{nx}}} \right)\\ &= Am.\frac{d}{{dx}}\left( {{e^{mx}}} \right) + Bn.\frac{d}{{dx}}\left( {{e^{nx}}} \right)\\ &= Am.{e^{mx}}.\frac{d}{{dx}}\left( {mx} \right) + Bn.{e^{nx}}.\frac{d}{{dx}}\left( {nx} \right)\\ &= A{m^2}{e^{mx}} + B{n^2}{e^{nx}}\end{align}

Thus,

\begin{align}\frac{{{d^2}y}}{{d{x^2}}} - \left( {m + n} \right)\frac{{dy}}{{dx}} + mny &= A{m^2}{e^{mx}} + B{n^2}{e^{nx}} - \left( {m + n} \right)\\&\qquad\left( {Am{e^{mx}} + Bn{e^{nx}}} \right) + mn\left( {A{e^{mx}} + B{e^{nx}}} \right)\\ &= A{m^2}{e^{mx}} + B{n^2}{e^{nx}} - A{m^2}{e^{mx}} - Bmn{e^{nx}} - Amn{e^{mx}} - B{n^2}{e^{nx}}\\&\qquad + Amn{e^{mx}} + Bmn{e^{nx}}\\ &= 0\end{align}

Hence proved.

## Chapter 5 Ex.5.7 Question 15

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

### Solution

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

Then,

\begin{align}\frac{{dy}}{{dx}} &= 500.\frac{d}{{dx}}\left( {{e^{7x}}} \right) + 600.\frac{d}{{dx}}\left( {{e^{ - 7x}}} \right)\\ &= 500.{e^{7x}}.\frac{d}{{dx}}\left( {7x} \right) + 600.{e^{ - 7x}}.\frac{d}{{dx}}\left( { - 7x} \right)\\& = 3500{e^{7x}} - 4200{e^{ - 7x}}\end{align}

Therefore,

\begin{align}\frac{{{d^2}y}}{{d{x^2}}}& = 3500{e^{7x}}.\frac{d}{{dx}}\left( {{e^{7x}}} \right) - 4200.\frac{d}{{dx}}\left( {{e^{ - 7x}}} \right)\\ &= 3500.{e^{7x}}.\frac{d}{{dx}}\left( {7x} \right) - 4200.{e^{ - 7x}}.\frac{d}{{dx}}\left( { - 7x} \right)\\ &= 7 \times 3500.{e^{7x}} + 7 \times 4200.{e^{ - 7x}}\\ &= 49 \times 500.{e^{7x}} + 49 \times 600{e^{ - 7x}}\\ &= 49\left( {500{e^{7x}} + 600{e^{ - 7x}}} \right)\\ &= 49y\end{align}

Hence proved.

## Chapter 5 Ex.5.7 Question 16

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

### Solution

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

\begin{align}& \Rightarrow \; {e^y}\left( {x + 1} \right) = 1\\& \Rightarrow \; {e^y} = \frac{1}{{x + 1}}\end{align}

Taking log on both sides, we get

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

Differentiating with respect to $$x$$, we get

\begin{align}\frac{{dy}}{{dx}}&= \left( {x + 1} \right)\frac{d}{{dx}}\left( {\frac{1}{{x + 1}}} \right)\\ &= \left( {x + 1} \right).\frac{{ - 1}}{{{{\left( {x + 1} \right)}^2}}}\\ &= \frac{{ - 1}}{{x + 1}}\end{align}

Therefore,

\begin{align}\frac{{{d^2}y}}{{d{x^2}}} &= \frac{d}{{dx}}\left( {\frac{{ - 1}}{{x + 1}}} \right) = - \left( {\frac{{ - 1}}{{{{\left( {x + 1} \right)}^2}}}} \right)\\ &= \frac{1}{{{{\left( {x + 1} \right)}^2}}} = {\left( {\frac{{ - 1}}{{x + 1}}} \right)^2}\\ &= {\left( {\frac{{dy}}{{dx}}} \right)^2}\end{align}

Hence proved.

## Chapter 5 Ex.5.7 Question 17

If $$y = {\left( {{{\tan }^{ - 1}}x} \right)^2}$$, show that $${\left( {{x^2} + 1} \right)^2}{y_2} + 2x\left( {{x^2} + 1} \right){y_1} = 2$$

### Solution

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

Then,

\begin{align} &\Rightarrow \; {y_1} = 2{\tan ^{ - 1}}x\frac{d}{{dx}}\left( {{{\tan }^{ - 1}}x} \right)\\ &\Rightarrow \; {y_1} = 2{\tan ^{ - 1}}x.\left( {\frac{1}{{1 + {x^2}}}} \right)\\ &\Rightarrow \; \left( {1 + {x^2}} \right){y_1} = 2{\tan ^{ - 1}}x\end{align}

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

\begin{align} &\Rightarrow \; \left( {1 + {x^2}} \right){y_2} + 2x{y_1} = 2\left( {\frac{1}{{1 + {x^2}}}} \right)\\ &\Rightarrow \; {\left( {1 + {x^2}} \right)^2}{y_2} + 2x\left( {1 + {x^2}} \right){y_1} = 2\end{align}

Hence proved.

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