Question

How do you differentiate `arctan(sinx)`?

Answer 1

`y' = 1/(1 + sin^2 x) cos x`

Explanation:

It is the derivative of a function arctan, of a function sin. It is equal to the derivative of the first function arctan, expressed in terms of its argument, multiplied by the derivative of the argument with respect to x.

The derivative of arctan u is `1/(1 + u^2)`. When u is `sin(x)` it is simply `1/(1 + sin^2 x)`. The derivative of `sin(x)` is `cosx`. Multiplying both yields

`y' = 1/(1 + sin^2 x) cos x`