How do I find the derivative of # y = (arcsin(x))^-1#?

1 Answer
Bill K.
Jun 26, 2015

#dy/dx=-(arcsin(x))^(-2)* 1/sqrt{1-x^2}=-1/(arcsin^{2}(x)sqrt{1-x^2})#

Explanation:

Use the Chain Rule: #d/dx(f(g(x)))=f'(g(x))*g'(x)# with #f(x)=x^{-1}# and #g(x)=arcsin(x)#. Since #f'(x)=-x^{-2}=-1/x^2# and #g'(x)=1/sqrt{1-x^2}#, it follows that

#d/dx((arcsin(x))^{-1})=-(arcsin(x))^(-2)* 1/sqrt{1-x^2}#

#=-1/(arcsin^{2}(x)sqrt{1-x^2})#

Related questions
See all questions in Differentiating Inverse Trigonometric Functions
Impact of this question
1344 views around the world
You can reuse this answer
Creative Commons License