Question

How do you find the derivative of the function: `sin(arccosx)`?

Answer 1

`d/dx(sin(arccosx)) = cos(arccosx)*d/dx(arccosx) = x * (-1)/sqrt(1-x^2)` `= (-x)/sqrt(1-x^2)`

Explanation:

It is possible to use the chain rule and the derivatives of `sin(u)` and `arccos(x)`, then use trigonometry to simplify. (As above.)
But I prefer to do the trigonometry first.

`sin(arccosx)` is the sine of a number in `[0,pi]` whose cosine is `x`. (Definition of `arccosx`)

The sine of such an angle is `sqrt(1-x^2)`.

That is,

`sin(arccosx) = sqrt(1-x^2)`

And `d/dx(sqrt(1-x^2)) = (-2x)/(2sqrt(1-x^2)) = (-x)/sqrt(1-x^2)`