Question

What's the derivative of `arctan(x)^(1/2)`?

Answer 1

so `\ y' = \frac{1}{2 \sqrt x} .\frac{1}{x+1 }`

Explanation:

let `\tan y = x^(1/2)`

so `\sec^2 y \ y' = \frac{1}{2 \sqrt x}`

using the `tan^2 + 1 = sec^2` identity, `sec^2 y = tan^2 y + 1 = x + 1`

so `\ y' = \frac{1}{2 \sqrt x} \frac{1}{\sec^2 y } = \frac{1}{2 \sqrt x} .\frac{1}{x+1 }`