NCERT Solutions for Exercise 5.7 Class 12 Maths Chapter 5 - Continuity and Differentiability

Question:2 Find the second order derivatives of the functions given in Exercises 1 to 10.

$x^{20}$

Answer:

Given function is
$y=x^{20}$
Now, differentiation w.r.t. x
$\frac{dy}{dx}=20x^{19}$
Now, the second-order derivative is
$\frac{d^{2}y}{d{x}^2}=20.19x^{18}=380x^{18}$
Therefore, second-order derivative is $\frac{d^{2}y}{d{x}^2}=380x^{18}$

Question:3 Find the second order derivatives of the functions given in Exercises 1 to 10.

$x\cos x$

Answer:

Given function is
$y=x\cos x$
Now, differentiation w.r.t. x
$\frac{dy}{dx}=\cos x+x(-\sin x)=\cos x-x\sin x$
Now, the second-order derivative is
$\frac{d^{2}y}{d{x}^2}=-\sin x-(\sin x+x\cos x)=-2\sin x-x\sin x$
Therefore, the second-order derivative is $\frac{d^{2}y}{d{x}^2}=-2\sin x-x\sin x$

Question:4 Find the second order derivatives of the functions given in Exercises 1 to 10.

$\log x$

Answer:

Given function is
$y=\log x$
Now, differentiation w.r.t. x
$\frac{dy}{dx}=\frac{1}{x}$
Now, second order derivative is
$\frac{d^{2}y}{d{x}^2}=\frac{-1}{x^2}$
Therefore, second order derivative is $\frac{d^{2}y}{d{x}^2}=\frac{-1}{x^2}$

Question:5 Find the second order derivatives of the functions given in Exercises 1 to 10.

$x^3\log x$

Answer:

Given function is
$y=x^3\log x$
Now, differentiation w.r.t. x
$\frac{dy}{dx}=3x^2.\log x+x^3.\frac{1}{x}=3x^2.\log x+x^2$
Now, the second-order derivative is
$\frac{d^{2}y}{d{x}^2}=6x.\log x+3x^2.\frac{1}{x}+2x=6x.\log x+3x+2x=x(6.\log x+5)$
Therefore, the second-order derivative is $\frac{d^{2}y}{d{x}^2}=x(6.\log x+5)$

Question:6 Find the second order derivatives of the functions given in Exercises 1 to 10.

$e^x\sin 5x$

Answer:

Given function is
$y=e^x\sin 5x$
Now, differentiation w.r.t. x
$\frac{dy}{dx}=e^x.\sin 5x+e^x.5\cos 5x=e^x(\sin 5x+5\cos 5x)$
Now, second order derivative is
$\frac{d^{2}y}{d{x}^2}=e^x(\sin 5x+5\cos 5x)+e^x(5\cos 5x+5.(-5\sin 5x))$
$=e^x(\sin 5x+5\cos 5x)+e^x(5\cos 5x-25\sin 5x)=e^x(10\cos 5x-24\sin 5x)$
$=2e^x(5\cos 5x-12\sin 5x)$
Therefore, second order derivative is $\frac{dy}{dx}=2e^x(5\cos 5x-12\sin 5x)$

Question:7 Find the second order derivatives of the functions given in Exercises 1 to 10.

$e^{6x}\cos 3x$

Answer:

Given function is
$y=e^{6x}\cos 3x$
Now, differentiation w.r.t. x
$\frac{dy}{dx}=6e^{6x}.\cos 3x+e^{6x}.(-3\sin 3x)=e^{6x}(6\cos 3x-3\sin 3x)$
Now, second order derivative is
$\frac{d^{2}y}{d{x}^2}=6e^{6x}(6\cos 3x-3\sin 3x)+e^{6x}(6.(-3\sin 3x)-3.3\cos 3x)$
$=6e^{6x}(6\cos 3x-3\sin 3x)-e^{6x}(18\sin 3x+9\cos 3x)$
$e^{6x}(27\cos 3x-36\sin 3x)=9e^{6x}(3\cos 3x-4\sin 3x)$
Therefore, second order derivative is $\frac{dy}{dx}=9e^{6x}(3\cos 3x-4\sin 3x)$

Question:8 Find the second order derivatives of the functions given in Exercises 1 to 10.

${\tan }^{-1}x$

Answer:

Given function is
$y={\tan }^{-1}x$
Now, differentiation w.r.t. x
$\frac{dy}{dx}=\frac{d({\tan }^{-1}x)}{dx}=\frac{1}{1+x^2}$
Now, second order derivative is
$\frac{d^{2}y}{d{x}^2}=\frac{-1}{(1+x^2{)}^2}.2x=\frac{-2x}{(1+x^2{)}^2}$
Therefore, second order derivative is $\frac{d^{2}y}{d{x}^2}=\frac{-2x}{(1+x^2{)}^2}$

Question:9 Find the second order derivatives of the functions given in Exercises 1 to 10.

$\log (\log x)$

Answer:

Given function is
$y=\log (\log x)$
Now, differentiation w.r.t. x
$\frac{dy}{dx}=\frac{d(\log (\log x))}{dx}=\frac{1}{\log x}.\frac{1}{x}=\frac{1}{x\log x}$
Now, second order derivative is
$\frac{d^{2}y}{d{x}^2}=\frac{-1}{(x\log x{)}^2}.(1.\log x+x.\frac{1}{x})=\frac{-(\log x+1)}{(x\log x{)}^2}$
Therefore, second order derivative is $\frac{d^{2}y}{d{x}^2}=\frac{-(\log x+1)}{(x\log x{)}^2}$

Question:10 Find the second order derivatives of the functions given in Exercises 1 to 10.

$\sin (\log x)$

Answer:

Given function is
$y=\sin (\log x)$
Now, differentiation w.r.t. x
$\frac{dy}{dx}=\frac{d(\sin (\log x))}{dx}=\cos (\log x).\frac{1}{x}=\frac{\cos (\log x)}{x}$
Now, second order derivative is
Using Quotient rule
$\frac{d^{2}y}{d{x}^2}=\frac{-\sin (\log x)\frac{1}{x}.x-\cos (\log x).1}{x^2}=\frac{-(\sin (\log x)+\cos (\log x))}{x^2}$
Therefore, second order derivative is $\frac{d^{2}y}{d{x}^2}=\frac{-(\sin (\log x)+\cos (\log x))}{x^2}$

Question:11 If $y=5\cos x-3\sin x$ prove that $\frac{d^{2}y}{d{x}^2}+y=0$

Answer:

Given function is
$y=5\cos x-3\sin x$
Now, differentiation w.r.t. x
$\frac{dy}{dx}=\frac{d(5\cos x-3\sin x)}{dx}=-5\sin x-3\cos x$
Now, the second-order derivative is
$\frac{d^{2}y}{d{x}^2}=\frac{d^2(-5\sin x-3\cos x)}{d{x}^2}=-5\cos x+3\sin x$
Now,
$\frac{d^{2}y}{d{x}^2}+y=-5\cos x+3\sin x+5\cos x-3\sin x=0$
Hence proved

Question:12 If $y={\cos }^{-1}x$ Find $\frac{d^{2}y}{d{x}^2}$ in terms of y alone.

Answer:

Given function is
$y={\cos }^{-1}x$
Now, differentiation w.r.t. x
$\frac{dy}{dx}=\frac{d({\cos }^{-1}x)}{dx}=\frac{-1}{\sqrt{1-x^2}}$
Now, second order derivative is
$\frac{d^{2}y}{d{x}^2}=\frac{d^2(\frac{-1}{\sqrt{1-x^2}})}{d{x}^2}=\frac{-(-1)}{(\sqrt{1-x^2}{)}^2}.(-2x)=\frac{-2x}{1-x^2}$ -(i)
Now, we want $\frac{d^{2}y}{d{x}^2}$ in terms of y
$y={\cos }^{-1}x$
$x=\cos y$
Now, put the value of x in (i)

$\frac{d^{2}y}{d{x}^2}=\frac{-2\cos y}{1-{\cos }^{2}y}=\frac{-2\cos y}{{\sin }^{2}y}=-2\cot y\mathrm{cosec}y$
$\frac{d^{2}y}{d{x}^2}=\frac{-2\cos y}{1-{\cos }^{2}y}=\frac{-2\cos y}{{\sin }^{2}y}=-2\cot y\mathrm{cosec}y$
$(\because 1-{\cos }^{2}x={\sin }^{2}x,\frac{\cos x}{\sin x}=\cot x,\text{and}\frac{1}{\sin x}=\mathrm{cosec}x)$
Therefore, answer is $\frac{d^{2}y}{d{x}^2}=-2\cot y\mathrm{cosec}y$

Question:13 If $y=3\cos (\log x)+4\sin (\log x)$, show that $x^2{y}_2+x{y}_1+y=0$

Answer:

Given function is
$y=3\cos (\log x)+4\sin (\log x)$
Now, differentiation w.r.t. x
$y_1=\frac{dy}{dx}=\frac{d(3\cos (\log x)+4\sin (\log x))}{dx}=-3\sin (\log x).\frac{1}{x}+4\cos (\log x).\frac{1}{x}$
$=\frac{4\cos (\log x)-3\sin (\log x)}{x}$ -(i)
Now, second order derivative is
By using the Quotient rule
$y_2=\frac{d^{2}y}{d{x}^2}=\frac{d^2\left(\frac{4\cos (\log x)-3\sin (\log x)}{x}\right)}{d{x}^2}$
$=\frac{(-4\sin (\log x)\cdot \frac{1}{x}-3\cos (\log x)\cdot \frac{1}{x})\cdot x-1\cdot (4\cos (\log x)-3\sin (\log x))}{x^2}$
$=\frac{-\sin (\log x)+7\cos (\log x)}{x^2}\text{-(ii)}$
Now, from equation (i) and (ii) we will get $y_{1}and{y}_2$
Now, we need to show
$x^2{y}_2+x{y}_1+y=0$
Put the value of $y_{1}and{y}_2$ from equation (i) and (ii)
$x^2\left(\frac{-\sin (\log x)+7\cos (\log x)}{x^2}\right)+x\left(\frac{4\cos (\log x)-3\sin (\log x)}{x}\right)+3\cos (\log x)+4\sin (\log x)$
$-\sin (\log x)-7\cos (\log x)+4\cos (\log x)-3\sin (\log x)+3\cos (\log x)+4\sin (\log x)$
$=0$
Hence proved

Question:14 If $y=A{e}^{mx}+B{e}^{nx}$ , show that $\frac{d^{2}y}{d{x}^2}-(m+n)\frac{dy}{dx}+mny=0$

Answer:

Given function is
$y=A{e}^{mx}+B{e}^{nx}$
Now, differentiation w.r.t. x
$\frac{dy}{dx}=\frac{d(A{e}^{mx}+B{e}^{nx})}{dx}=mA{e}^{mx}+nB{e}^{nx}$ -(i)
Now, second order derivative is
$\frac{d^{2}y}{d{x}^2}=\frac{d^2(mA{e}^{mx}+nB{e}^{nx})}{d{x}^2}=m^{2}A{e}^{mx}+n^{2}B{e}^{nx}$ -(ii)
Now, we need to show
$\frac{d^{2}y}{d{x}^2}-(m+n)\frac{dy}{dx}+mny=0$
Put the value of $\frac{d^{2}y}{d{x}^2}and\frac{dy}{dx}$ from equation (i) and (ii)
$m^{2}A{e}^{mx}+n^{2}B{e}^{nx}-(m+n)(mA{e}^{mx}+nB{x}^{nx})+mn(A{e}^{mx}+B{e}^{nx})$
$m^{2}A{e}^{mx}+n^{2}B{e}^{nx}-m^{2}A{e}^{mx}-mnB{x}^{nx}-mnA{e}^{mx}-n^{2}B{e}^{nx}+mnA{e}^{mx}$$+mnB{e}^{nx}$
$=0$
Hence proved

Question:15 If $y=500e^{7x}+600e^{-7x}$ , show that $\frac{d^{2}y}{d{x}^2}=49y$
Answer:

Given function is
$y=500e^{7x}+600e^{-7x}$
Now, differentiation w.r.t. x
$\frac{dy}{dx}=\frac{d(500e^{7x}+600e^{-7x})}{dx}=7.500e^{7x}-7.600e^{-7x}=3500e^{7x}-4200e^{-7x}$ -(i)
Now, second order derivative is
$\frac{d^{2}y}{d{x}^2}=\frac{d^2(3500e^{7x}-4200e^{-7x})}{d{x}^2}$
$=7.3500e^{7x}-(-7).4200e^{-7x}=24500e^{7x}+29400e^{-7x}$ -(ii)
Now, we need to show
$\frac{d^{2}y}{d{x}^2}=49y$
Put the value of $\frac{d^{2}y}{d{x}^2}$ from equation (ii)
$24500e^{7x}+29400e^{-7x}=49(500e^{7x}+600e^{-7x})$
$=24500e^{7x}+29400e^{-7x}$
Hence, L.H.S. = R.H.S.
Hence proved

Question:16 If $e^y(x+1)=1$ show that $\frac{d^{2}y}{d{x}^2}=(\frac{dy}{dx}{)}^2$

Answer:

Given function is
$e^y(x+1)=1$
We can rewrite it as
$e^y=\frac{1}{x+1}$
Now, differentiation w.r.t. x
$$\begin{gathered} \frac{d(e^y)}{dx}=\frac{d(\frac{1}{x+1})}{dx} \\ {e}^y.\frac{dy}{dx}=\frac{-1}{(x+1{)}^2} \\ \frac{1}{x+1}.\frac{dy}{dx}=\frac{-1}{(x+1{)}^2}(\because e^y=\frac{1}{x+1}) \\ \frac{dy}{dx}=\frac{-1}{x+1} \end{gathered}$$ -(i)
Now, second order derivative is
$\frac{d^{2}y}{d{x}^2}=\frac{d^2(\frac{-1}{x+1})}{d{x}^2}=\frac{-(-1)}{(x+1{)}^2}=\frac{1}{(x+1{)}^2}$ -(ii)
Now, we need to show
$\frac{d^{2}y}{d{x}^2}=(\frac{dy}{dx}{)}^2$
Put value of $\frac{d^{2}y}{d{x}^2}and\frac{dy}{dx}$ from equation (i) and (ii)
$\frac{1}{(x+1{)}^2}={\left(\frac{-1}{x+1}\right)}^2$
$=\frac{1}{(x+1{)}^2}$
Hence, L.H.S. = R.H.S.
Hence proved

Question:17 If $y=({\tan }^{-1}x{)}^2$ show that $(x^2+1{)}^2{y}_2+2x(x^2+1)y_1=2$

Answer:

Given function is
$y=({\tan }^{-1}x{)}^2$
Now, differentiation w.r.t. x
$y_1=\frac{dy}{dx}=\frac{d(({\tan }^{-1}x{)}^2)}{dx}=2.{\tan }^{-1}x.\frac{1}{1+x^2}=\frac{2{\tan }^{-1}x}{1+x^2}$ -(i)
Now, the second-order derivative is
By using the quotient rule
$y_2=\frac{d^{2}y}{d{x}^2}=\frac{d^2(\frac{2{\tan }^{-1}x}{1+x^2})}{d{x}^2}=\frac{2.\frac{1}{1+x^2}.(1+x^2)-2{\tan }^{-1}x(2x)}{(1+x^2{)}^2}=\frac{2-4x{\tan }^{-1}x}{(1+x^2{)}^2}$ -(ii)
Now, we need to show
$(x^2+1{)}^2{y}_2+2x(x^2+1)y_1=2$
Put the value from equation (i) and (ii)
$$\begin{gathered} (x^2+1{)}^2.\frac{2-4x{\tan }^{-1}x}{(1+x^2{)}^2}+2x(x^2+1).\frac{2{\tan }^{-1}x}{x^2+1} \\ \Rightarrow 2-4x{\tan }^{-1}x+4x{\tan }^{-1}x=2 \end{gathered}$$
Hence, L.H.S. = R.H.S.
Hence proved