Exercise 5.7 (Revised) - Chapter 5 - Continuity & Differentiability - Ncert Solutions class 12 - Maths

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Chapter 5: Continuity & Differentiability NCERT Solutions Class 12 Maths

Ex 5.7 Question 1:

Find the second order derivatives of the function.
$
x^2+3 x+2
$

Answer :
Let $y=x^2+3 x+2$
Then,
$
\begin{aligned}
& \frac{d y}{d x}=\frac{d}{d x}\left(x^2\right)+\frac{d}{d x}(3 x)+\frac{d}{d x}(2)=2 x+3+0=2 x+3 \\
& \therefore \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}
$

Ex 5.7 Question 2:

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

Answer :
Let $y=x^{20}$
Then,
$
\begin{aligned}
& \frac{d y}{d x}=\frac{d}{d x}\left(x^{20}\right)=20 x^{19} \\
& \therefore \frac{d^2 y}{d x^2}=\frac{d}{d x}\left(20 x^{19}\right)=20 \frac{d}{d x}\left(x^{19}\right)=20 \cdot 19 \cdot x^{18}=380 x^{18}
\end{aligned}
$

Ex 5.7 Question 3 :

Find the second order derivatives of the function.
$
x \cdot \cos x
$

Answer :
Let $y=x \cdot \cos x$
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 \\
& \therefore \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)\right] \\
& =-\sin x-(\sin x+x \cos x) \\
& =-(x \cos x+2 \sin x) \\
&
\end{aligned}
$

Ex 5.7 Question 4:

Find the second order derivatives of the function.
$\log x$

Answer:
Let $y=\log x$
Then,
$
\begin{aligned}
& \frac{d y}{d x}=\frac{d}{d x}(\log x)=\frac{1}{x} \\
& \therefore \frac{d^2 y}{d x^2}=\frac{d}{d x}\left(\frac{1}{x}\right)=\frac{-1}{x^2}
\end{aligned}
$

Ex 5.7 Question 5 :

Find the second order derivatives of the function.
$
x^3 \log x
$

Answer:
Let $y=x^3 \log x$
Then,
$
\begin{aligned}
\frac{d y}{d x}= & \frac{d}{d x}\left[x^3 \log x\right]=\log x \cdot \frac{d}{d x}\left(x^3\right)+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) \\
\therefore \frac{d^2 y}{d x^2} & =\frac{d}{d x}\left[x^2(1+3 \log x)\right] \\
& =(1+3 \log x) \cdot \frac{d}{d x}\left(x^2\right)+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 x \log x+3 x \\
& =5 x+6 x \log x \\
& =x(5+6 \log x)
\end{aligned}
$

Ex 5.7 Question 6:

Find the second order derivatives of the function.
$
e^x \sin 5 x
$

Answer :
Let $y=e^x \sin 5 x$
$
\begin{aligned}
& \frac{d y}{d x}=\frac{d}{d x}\left(e^x \sin 5 x\right)=\sin 5 x \cdot \frac{d}{d x}\left(e^x\right)+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) \\
& \text { Then, } \therefore \frac{d^2 y}{d x^2}=\frac{d}{d x}\left[e^x(\sin 5 x+5 \cos 5 x)\right] \\
& =(\sin 5 x+5 \cos 5 x) \cdot \frac{d}{d x}\left(e^x\right)+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)\right] \\
& =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}
$

Ex 5.7 Question 7 :

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

Answer:
Let $y=e^{6 x} \cos 3 x$
Then,
$
\begin{aligned}
\frac{d y}{d x} & =\frac{d}{d x}\left(e^{6 x} \cdot \cos 3 x\right)=\cos 3 x \cdot \frac{d}{d x}\left(e^{6 x}\right)+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 \quad \ldots(1) \\
\therefore \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\right)=6 \cdot \frac{d}{d x}\left(e^{6 x} \cos 3 x\right)-3 \cdot \frac{d}{d x}\left(e^{6 x} \sin 3 x\right) \\
& =6 \cdot\left[6 e^{6 x} \cos 3 x-3 e^{6 x} \sin 3 x\right]-3 \cdot\left[\sin 3 x \cdot \frac{d}{d x}\left(e^{6 x}\right)+e^{6 x} \cdot \frac{d}{d x}(\sin 3 x)\right] \quad \text { [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\right] \\
& =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}
$

Ex 5.7 Question 8 :

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

Answer:
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} \\
& \begin{aligned}
\therefore \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}
\end{aligned}
$

Ex 5.7 Question 9 :

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

Answer :
Let $y=\log (\log x)$
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}{x \log x}=(x \log x)^{-1} \\
& \begin{aligned}
\therefore \frac{d^2 y}{d x^2} & =\frac{d}{d x}\left[(x \log x)^{-1}\right]=(-1) \cdot(x \log x)^{-2} \cdot \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)\right] \\
& =\frac{-1}{(x \log x)^2} \cdot\left[\log x \cdot 1+x \cdot \frac{1}{x}\right]=\frac{-(1+\log x)}{(x \log x)^2}
\end{aligned}
\end{aligned}
$

Ex 5.7 Question 10:

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

Answer:
Let $y=\sin (\log x)$
Then,
$
\begin{aligned}
& \frac{d y}{d x}=\frac{d}{d x}[\sin (\log x)]=\cos (\log x) \cdot \frac{d}{d x}(\log x)=\frac{\cos (\log x)}{x} \\
& \therefore \frac{d^2 y}{d x^2}=\frac{d}{d x}\left[\frac{\cos (\log x)}{x}\right] \\
& =\frac{x \cdot \frac{d}{d x}[\cos (\log x)]-\cos (\log x) \cdot \frac{d}{d x}(x)}{x^2} \\
& =\frac{x \cdot\left[-\sin (\log x) \cdot \frac{d}{d x}(\log x)\right]-\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}
$

Ex 5.7 Question 11:

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

Answer:
It is given that, $y=5 \cos x-3 \sin x$
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) \\
& \therefore \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)\right] \\
& =-[5 \cos x+3(-\sin x)] \\
& =-[5 \cos x-3 \sin x] \\
& =-y \\
& \therefore \frac{d^2 y}{d x^2}+y=0 \\
&
\end{aligned}
$

Hence, proved.

Ex 5.7 Question 12:

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

Answer :
It is given that, $y=\cos ^{-1} x$
Then,
$
\begin{aligned}
& \frac{d y}{d x}=\frac{d}{d x}\left(\cos ^{-1} x\right)=\frac{-1}{\sqrt{1-x^2}}=-\left(1-x^2\right)^{\frac{-1}{2}} \\
& \begin{aligned}
\frac{d^2 y}{d x^2} & =\frac{d}{d x}\left[-\left(1-x^2\right)^{\frac{-1}{2}}\right] \\
& =-\left(-\frac{1}{2}\right) \cdot\left(1-x^2\right)^{\frac{-3}{2}} \cdot \frac{d}{d x}\left(1-x^2\right) \\
& =\frac{1}{2 \sqrt{\left(1-x^2\right)^3}} \times(-2 x)
\end{aligned} \\
& \Rightarrow \frac{d^2 y}{d x^2}=\frac{-x}{\sqrt{\left(1-x^2\right)^3}} \\
& y=\cos ^{-1} x \Rightarrow x=\cos y
\end{aligned}
$

Putting $x=\cos y$ in equation (i), we obtain
$
\begin{aligned}
& \frac{d^2 y}{d x^2}=\frac{-\cos y}{\sqrt{\left(1-\cos ^2 y\right)^3}} \\
& \Rightarrow \frac{d^2 y}{d x^2}=\frac{-\cos y}{\sqrt{\left(\sin ^2 y\right)^3}}
\end{aligned}
$

$\begin{aligned}
& =\frac{-\cos y}{\sin ^3 y} \\
& =\frac{-\cos y}{\sin y} \times \frac{1}{\sin ^2 y} \\
\Rightarrow \frac{d^2 y}{d x^2} & =-\cot y \cdot \operatorname{cosec}^2 y
\end{aligned}$

Ex 5.7 Question 13:

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

Answer :
It is given that, $y=3 \cos (\log x)+4 \sin (\log x)$
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)\right]+4 \cdot\left[\cos (\log x) \cdot \frac{d}{d x}(\log x)\right] \\
\therefore y_1 & =\frac{-3 \sin (\log x)}{x}+\frac{4 \cos (\log x)}{x}=\frac{4 \cos (\log x)-3 \sin (\log x)}{x} \\
\therefore y_2 & =\frac{d}{d x}\left(\frac{4 \cos (\log x)-3 \sin (\log x)}{x}\right) \\
& =\frac{x\{4 \cos (\log x)-3 \sin (\log x)\}^{\prime}-\{4 \cos (\log x)-3 \sin (\log x)\}(x)^{\prime}}{x^2} \\
& =\frac{x\left[4\{\cos (\log x)\}^{\prime}-3\{\sin (\log x)\}^{\prime}\right]-\{4 \cos (\log x)-3 \sin (\log x)\} \cdot 1}{x^2} \\
& =\frac{x\left[-4 \sin (\log x) \cdot(\log x)^{\prime}-3 \cos (\log x) \cdot(\log x)^{\prime}\right]-4 \cos (\log x)+3 \sin (\log x)}{x^2}
\end{aligned}
$

$
\begin{aligned}
& =\frac{x\left[-4 \sin (\log x) \cdot \frac{1}{x}-3 \cos (\log x) \cdot \frac{1}{x}\right]^{x^2}-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} \\
\therefore & x^2 y_2+x y_1+y \\
= & 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
\end{aligned}
$

Hence, proved.

Ex 5.7 Question 14 :

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

Answer:
It is given that, $y=A e^{m x}+B e^{n x}$
Then,

$\begin{aligned}
\frac{d y}{d x} & =A \cdot \frac{d}{d x}\left(e^{m x}\right)+B \cdot \frac{d}{d x}\left(e^{n x}\right)=A \cdot e^{m x} \cdot \frac{d}{d x}(m x)+B \cdot e^{n x} \cdot \frac{d}{d x}(n x)=A m e^{n x}+B n e^{n x} \\
\frac{d^2 y}{d x^2} & =\frac{d}{d x}\left(A m e^{m x}+B n e^{n x}\right)=A m \cdot \frac{d}{d x}\left(e^{m x}\right)+B n \cdot \frac{d}{d x}\left(e^{n x}\right) \\
& =A m \cdot e^{m x} \cdot \frac{d}{d x}(m x)+B n \cdot e^{n x} \cdot \frac{d}{d x}(n x)=A m^2 e^{m x}+B n^2 e^{n x}
\end{aligned}$

$\therefore \frac{d^2 y}{d x^2}-(m+n) \frac{d y}{d x}+m n y$

$
\begin{aligned}
& =A m^2 e^{m x}+B n^2 e^{n x}-(m+n) \cdot\left(A m e^{n x}+B n e^{n x}\right)+m n\left(A e^{m x}+B e^{n x}\right) \\
& =A m^2 e^{m x}+B n^2 e^{n x}-A m^2 e^{n 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.

Ex 5.7 Question 15:

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

Answer :
It is given that, $y=500 e^{7 x}+600 e^{-7 x}$
Then,
$
\begin{aligned}
\frac{d y}{d x}= & 500 \cdot \frac{d}{d x}\left(e^{7 x}\right)+600 \cdot \frac{d}{d x}\left(e^{-7 x}\right) \\
= & 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} \\
\therefore \frac{d^2 y}{d x^2} & =3500 \cdot \frac{d}{d x}\left(e^{7 x}\right)-4200 \cdot \frac{d}{d x}\left(e^{-7 x}\right) \\
& =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 \times 3500 \cdot e^{7 x}+7 \times 4200 \cdot e^{-7 x} \\
& =49 \times 500 e^{7 x}+49 \times 600 e^{-7 x} \\
& =49\left(500 e^{7 x}+600 e^{-7 x}\right) \\
& =49 y
\end{aligned}
$

Hence, proved.

Ex 5.7 Question 16:

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

Answer:
The given relationship is $e^y(x+1)=1$
$
\begin{aligned}
& e^y(x+1)=1 \\
& \Rightarrow e^y=\frac{1}{x+1}
\end{aligned}
$

Taking logarithm on both the sides, we obtain
$
y=\log \frac{1}{(x+1)}
$

Differentiating this relationship with respect to $x$, we obtain
$
\begin{aligned}
& \frac{d y}{d x}=(x+1) \frac{d}{d x}\left(\frac{1}{x+1}\right)=(x+1) \cdot \frac{-1}{(x+1)^2}=\frac{-1}{x+1} \\
& \therefore \frac{d^2 y}{d x^2}=-\frac{d}{d x}\left(\frac{1}{x+1}\right)=-\left(\frac{-1}{(x+1)^2}\right)=\frac{1}{(x+1)^2} \\
& \Rightarrow \frac{d^2 y}{d x^2}=\left(\frac{-1}{x+1}\right)^2 \\
& \Rightarrow \frac{d^2 y}{d x^2}=\left(\frac{d y}{d x}\right)^2
\end{aligned}
$

Hence, proved.

Ex 5.7 Question 17:

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

Answer :
The given relationship is $y=\left(\tan ^{-1} x\right)^2$
Then,
$
\begin{aligned}
& y_1=2 \tan ^{-1} x \frac{d}{d x}\left(\tan ^{-1} x\right) \\
& \Rightarrow y_1=2 \tan ^{-1} x \cdot \frac{1}{1+x^2} \\
& \Rightarrow\left(1+x^2\right) y_1=2 \tan ^{-1} x
\end{aligned}
$

Again differentiating with respect to $x$ on both the sides, we obtain
$
\begin{aligned}
& \left(1+x^2\right) y_2+2 x y_1=2\left(\frac{1}{1+x^2}\right) \\
& \Rightarrow\left(1+x^2\right)^2 y_2+2 x\left(1+x^2\right) y_1=2
\end{aligned}
$

Hence, proved.