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#### Explain solution RD Sharma class 12 Chapter 11 Higher Order Derivatives Exercise Multiple Choice Question question 17

(c)

Hint:

We must have known about the derivative of   $\sin x,\cos x$  and  $\tan x$ .

Given:

$y=e^{\tan x} \$

Explanation:

$y=e^{\tan x} \$

$y_{1}=\frac{d y}{d x}=e^{\tan x} \times \sec ^{2} x \\$

\begin{aligned} & &\frac{d y}{d x}=e^{\tan x}\left(\sec ^{2} x\right) \end{aligned}

Again differentiate with respect to $x$ ,

$\frac{d^{2} y}{d x^{2}}=\left[e^{\tan x}\left(\sec ^{2} x\right)\left(\sec ^{2} x\right)+e^{\tan x}\left(2 \sec ^{2} x \tan x\right)\right] \$

\begin{aligned} &\ &\frac{d^{2} y}{d x^{2}}=e^{\tan x}\left[\sec ^{4} x+2 \sec ^{2} x \tan x\right] \end{aligned}

$\frac{d^{2} y}{d x^{2}}=e^{\tan x} \cdot \sec ^{2} x\left[\sec ^{2} x+2 \tan x\right] \\$

$\frac{d^{2} y}{d x^{2}}=e^{\tan x} \cdot \sec ^{2} x\left[\frac{1}{\cos ^{2} x}+2 \frac{\sin x}{\cos x}\right]$

\begin{aligned} & \\ &\frac{d^{2} y}{d x^{2}}=e^{\operatorname{lan} x} \cdot \sec ^{2} x\left[\frac{1+2 \sin x \cos x}{\cos ^{2} x}\right] \end{aligned}

$\cos ^{2} x \frac{d^{2} y}{d x^{2}}=y_{1}[1+2 \sin x \cos x]$

\begin{aligned} &\\ &\cos ^{2} x \frac{d^{2} y}{d x^{2}}=y_{1}[1+\sin 2 x] \end{aligned}                                        $[\because 2 \sin x \cos x=\sin 2 x]$