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Need solution for RD Sharma maths class 12 chapter Higher Order Derivatives exercise 11.1 question 1 subquestion (vii)

Answers (1)

Answer:

        \frac{-2x}{(1+x^{2})^{2}}

Hint:

You must know about derivative of tan-1x

Given:

        tan-1x

Solution:

Let\: \: y=tan^{-1}x

        \begin{aligned} &\frac{d y}{d x}=\frac{1}{1+x^{2}} \quad\left(\frac{d \tan ^{-1} x}{d x}=\frac{1}{1+x^{2}}\right) \\ & \end{aligned}

Use Quotient rule

        \begin{aligned} &\text { As } \frac{u}{v}=\frac{u^{1} v-v^{1} u}{v^{2}} \\ &\text { Where } v=1+x^{2} \& u=1 \\ &\frac{d^{2} y}{d x^{2}}=\frac{\left(1+x^{2}\right) \frac{d}{d x} 1-1 \cdot \frac{d}{d x}\left(1+x^{2}\right)}{\left(1+x^{2}\right)^{2}} \\ &\left.\frac{d^{2} y}{d x^{2}}=\frac{0\left(1+x^{2}\right)-1.2 x}{\left(1+x^{2}\right)^{2}} \quad \text { ( } \frac{d}{d x} 1=0, \frac{d}{d x}\left(1+x^{2}\right)=2 x\right) \\ &\frac{d^{2} y}{d x^{2}}=\frac{0-2 x}{\left(1+x^{2}\right)^{2}} \\ &\frac{d^{2} y}{d x^{2}}=\frac{-2 x}{\left(1+x^{2}\right)^{2}} \end{aligned}

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