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

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

Answers (1)

Answer:

                (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]

 

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