the derivative of acrcos?

Question

I'm just looking for the derivative of arccos

arctan- 1/(1+x^2)
arcsin- 1/sqrt(1-x^2)

Answer 1

You can always derive it if you forget. Set y = arccosx. We wish to find dy/dx. Recalling a property of inverse functions, we can rearrange y = arccosx as x = cosy. Now we can differentiate both sides with respect to x and get

1 = -(siny)*(dy/dx)
dy/dx = -1 / siny.

Now we just need to find siny. Well we know that cosy = x. I can think of two ways to finish. One is to construct a right triangle with one of its acute angles being y such that cosy = x. One such triangle has adjacent leg equal to x, the hypotenuse equal to 1, and the opposite leg equal to sqrt(1-x^2). Then siny = sqrt(1-x^2). We can plug this back in to our eqn for dy/dx above.

Another way is to use the identity (sinx)^2 + (cosx)^2 = 1 and solve for sine. This gives us the same result.

I find this useful cause I never remember the inverse trig derivatives.

Answer 2

Derivative of arccos is just negative of derivative of arcsin:

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

Answer 3

derivative of arccos is -1/sqrt(1-x^2)

Answer 4

arccos x = π/2 - arcsin x.
So the derivative of arccos x is -1/ √(1-x²).