Question

How do you find the derivative of the function: `y = arctan(cosx)`?

Answer 1

`(dy)/(dx)=(-sinx)/(1+cos^2x)`

Explanation:

WE are given `y=arctan(cosx)`

Let `y=f(x)=arctanx` and `g(x)=cosx`

Then `(dg(x))/(dx)=-sinx` and `y=arctan(g(x))`

and according to chain rule `(dy)/(dx)=(dy)/(dg(x))*(dg(x))/(dx)`

Now as `y=arctan(g(x))`, we have `(dy)/(dg(x))=1/(1+(g(x))^2)`

and as `g(x)=cosx`, `(dg)/(dx)=-sinx`

Hence `(dy)/(dx)=1/(1+(g(x))^2)*(-sinx)`

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