Question

What is the derivative of #y=arccos(x^2 )#?

Answer 1

By Chain Rule, we can find
`y'=-{2x}/{sqrt{1-x^4}}`.

Recall the chain rule `[f(g(x))]'=f'(g(x))cdot g'(x)`, and
`(arccos x)'=-1/sqrt{1-x^2}`.

Now, let us take the derivative of the posted function.
By Chain Rule,
`y'=-1/sqrt{1-(x^2)^2}cdot2x=-{2x}/sqrt{1-x^4}`