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

1 Answer
Jul 31, 2014

The answer is:

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

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

1.) #y = arccosx#

2.) #cos y = cos(arccosx)#

3.) #cos y = x#

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

4.) #-siny dy/dx = 1#

Solve for #dy/dx#:

5.) #dy/dx = -1/siny#

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

6.) #dy/dx = -1/sin(arccosx)#

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[arcsin x]# and #d/dx[arctan x]#, since they appear quite frequently in advanced differentiation problems.