Get Answers to all your Questions

header-bg qa

Please solve RD Sharma class 12 chapter 11 Higher Order Derivatives exercise very short answer type question 9 maths textbook solution

Answers (1)

Answer:

        \frac{d^{2} y}{d x^{2}}= \begin{cases}\frac{1}{x^{2}} & \text { if } 0<x<1 \\ \frac{-1}{x^{2}} & \text { if } x>1\end{cases}

Hint:

        \text { if } y=|p| \text { then } y=\left\{\begin{array}{c} p \text { if } 0<\mathrm{p}<1 \\ -p\; p<0 \text { or } p>1 \end{array}\right.

Given:

        y=|log\: ex|

Explanation:

 It is given that

        y=|log\: ex|

This can be written as,

        y=\left\{\begin{array}{cc} -\log e^{x} & \text { if } 0<x<1 \\ \log e^{x} & \text { if } x>1 \end{array}\right.

Diff y w.r.to x

        \frac{d y}{d x}=\left\{\begin{array}{lc} -1 / x & \text { if } 0<x<1 \\ 1 / x & \text { if } x>1 \end{array}\right.

Again diff w.r.to x

        \frac{d^{2} y}{d x^{2}}= \begin{cases}\frac{1}{x^{2}} & \text { if } 0<x<1 \\ \frac{-1}{x^{2}} & \text { if } x>1\end{cases}

Thus

        \frac{d^{2} y}{d x^{2}}= \begin{cases}\frac{1}{x^{2}} & \text { if } 0<x<1 \\ \frac{-1}{x^{2}} & \text { if } x>1\end{cases}

Posted by

Gurleen Kaur

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads