Question

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

Answer 1

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.