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

1 Answer
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.