Get Answers to all your Questions

header-bg qa

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

Answers (1)




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


                y=e^{\tan x} \


                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]


Posted by


View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support