Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • explain solution RD Sharma class 12 chapter Differentiation exercise 10.2 question 47 maths

explain solution RD Sharma class 12 chapter Differentiation exercise 10.2 question 47 maths

Answers (1)

Answer: \frac{16 x^{3}\left(\sin ^{-1} x^{4}\right)^{3}}{\sqrt{1-x^{8}}}

Hint: you must know the rules of solving derivative of inverse trigonometric function

Given: \left(\sin ^{-1} x^{4}\right)^{4}

Solution:

Let  y=\left(\sin ^{-1} x^{4}\right)^{4}

Differentiate with respect to x

\frac{d y}{d x}=\frac{d}{d x}\left(\sin ^{-1} x^{4}\right)^{4}

\begin{aligned} &=4\left(\sin ^{-1} x^{4}\right)^{3} \frac{\mathrm{d}}{\mathrm{dx}}\left(\sin ^{-1} x^{4}\right) \\\\ &=4\left(\sin ^{-1} x^{4}\right)^{3} \frac{1}{\sqrt{1-\left(x^{4}\right)^{2}}} \frac{\mathrm{d}}{\mathrm{dx}}\left(x^{4}\right) \end{aligned}

\begin{aligned} &=4\left(\sin ^{-1} x^{4}\right)^{3} \frac{4 x^{3}}{\sqrt{1-x^{8}}} \\\\ &\frac{d y}{d x} \Rightarrow \frac{16 x^{3}\left(\sin ^{-1} x^{4}\right)^{3}}{\sqrt{1-x^{8}}} \end{aligned}

 

Posted by

infoexpert26

View full answer

Crack CUET with india's "Best Teachers"

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

Similar Questions

Latest Question