Question

How do you find the derivative of `sqrt[arctan(x)]`?

Answer 1

`1/(2(x^2+1)sqrtarctan(x))`

Explanation:

`y=sqrtarctan(x)`

`y=(arctan(x))^(1/2)`

We will use the chain rule. The specific adaptation of the chain rule that we will be using here is where we use the power rule on the outside function:

`d/dxu^(1/2)=1/2u^(-1/2)(du)/dx`

So:

`dy/dx=1/2(arctan(x))^(-1/2)d/dxarctan(x)`

`dy/dx=1/(2sqrtarctan(x))d/dxarctan(x)`

Note that `d/dxarctan(x)=1/(x^2+1)`:

`dy/dx=1/(2(x^2+1)sqrtarctan(x))`

Answer 2

`1/(2(x^2+1)sqrtarctan(x))`

Explanation:

`y=sqrtarctan(x)`

`y^2=arctan(x)`

Differentiate both sides. The chain rule will be used on the left!

`2ydy/dx=1/(x^2+1)`

Since `2y=2sqrtarctan(x)`:

`dy/dx=1/(x^2+1)*1/(2sqrtarctan(x))`

`dy/dx=1/(2(x^2+1)sqrtarctan(x))`