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Please solve RD Sharma class 12 chapter Higher Order Derivatives exercise 11.1 question 1 subquestion (i) maths textbook solution

Answers (1)

Answer:

        6x+2sec^{2}x\: tan\, x

Hint:

        You must know about derivative of  tan x

Given:

        x^{3}+tan\: x

Solution:

Let\: \: y=x^{3}+tan\: x

        \begin{aligned} &\frac{d y}{d x}=3 x^{2}+\sec ^{2} x \quad \quad\left[\frac{d(\tan x)}{d x}=\sec ^{2} x \text { and } \frac{d x^{3}}{d x}=3 x^{2}\right] \\ &\frac{d^{2} y}{d x^{2}}=6 x+2 \sec x \cdot \sec x \tan x \frac{d \sec x}{d x}=\sec x \tan x \\ &\frac{d^{2} y}{d x^{2}}=6 x+2 \sec ^{2} x \tan x \end{aligned}

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Gurleen Kaur

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