Question

What is the derivative of `sin(arc cosx)`?

Answer 1

`-x/sqrt(1-x^2)`

Explanation:

`y=sin(arccosx)`

Method 1
(Use if you don't remember the derivative of `arccosx`., but remember your trigonometry.)

`sin(arccosx)` is the sine of a number (an angle) in `[0, pi]` whose cosine is `x`.
In `[0, pi]` the sine is positive, so: `sin(arccosx) = sqrt(1-x^2)`

Now `y=sin(arccosx) = sqrt(1-x^2)`, so

`y' = 1/(2sqrt(1-x^2)) (-2x) = -x/sqrt(1-x^2)`

Method 2:
(Use if you remember the derivative of `arccosx`.)
Use the chain rule. (And the fact that `cos(arccosx)=x`)

`y'=cos(arccosx) * d/dx(arccosx)`

`= x* (-1/sqrt(1-x^2)) = -x/sqrt(1-x^2)`