How do you find the derivative of # (arcsin x)^2#?

1 Answer
Shwetank Mauria
Nov 18, 2017

Derivative is #(2arcsinx)/sqrt(1-x^2)#

Explanation:

Let us first workout derivative of #arcsinx#. Let #y=arcsinx# i.e. #siny=x# and hence differentiating

#cosy*(dy)/(dx)=1# or #(dy)/(dx)=1/cosy=1/sqrt(1-sin^2y)=1/sqrt(1-x^2)#

Hence #d/(dx)(arcsinx)^2=2arcsinx xx d/(dx)arcsinx#

= #(2arcsinx)/sqrt(1-x^2)#

Related questions
See all questions in Differentiating Inverse Trigonometric Functions
Impact of this question
7171 views around the world
You can reuse this answer
Creative Commons License