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

Question

1308 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-02-25T17:15:21

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

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

Here we need to apply the chain rule successively:

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

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

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

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