Question

How do you find the derivatives of inverse function `f(x) = arctan(cos(3x))`?

Answer 1

`d/(dx)[arctan(u(v(x)))] = 1/(1+u^2)*((du)/(dv))((dv)/(dx))`

`d/(dx)[cosu] = -sinu*((du)/(dx))`

With `u(v) = cosv` and `v(x) = 3x`:

`d/(dx)[arctan(cos3x)] = 1/(1+cos^2(3x))*-sin(3x)*3`

`= (-3sin3x)/(1+cos^2(3x))`