Question

What is the derivative of `ln(x^2+1)^(1/2)`?

Answer 1

`f'(x)=(1)/((x^2+1))`

Explanation:

`f(x)=ln[(x^2+1)^(1/2)]`

Here we need to apply the chain rule successively:

`f'(x)=(1)/((x^2+1)^(1/2))xx[1/cancel(2)(x^2+1)^(-1/2)xxcancel(2)]`

`f'(x)=(1)/((x^2+1)^(1/2)xx(x^2+1)^(1/2))`

`:.f'(x)=(1)/((x^2+1))`