What is the derivative of y=arcsin(x)?

1 Answer
Jacob F.
Aug 3, 2014

The answer is:

dy/dx = 1/(sqrt(1-x^2))

This identity can be proven easily by applying sin to both sides of the original equation:

1.) y = arcsinx

2.) sin y = sin(arcsinx)

3.) sin y = x

We continue by using implicit differentiation, keeping in mind to use the chain rule on siny:

4.) cosy dy/dx = 1

Solve for dy/dx:

5.) dy/dx = 1/cosy

Now, substitution with our original equation yields dy/dx in terms of x:

6.) dy/dx = 1/cos(arcsinx)

At first this might not look all that great, but it can be simplified if one recalls the identity
sin(arccosx) = cos(arcsinx) = sqrt(1 - x^2).

7.) dy/dx = 1/sqrt(1 - x^2)

This is a good definition to memorize, along with d/dx[arccos x] and d/dx[arctan x], since they appear quite frequently in differentiation problems.

Related questions
See all questions in Differentiating Inverse Trigonometric Functions
Impact of this question
13933 views around the world
You can reuse this answer
Creative Commons License