Question

How do you find the derivative of `f(x)=ln(2x^2+1)`?

Answer 1

`(4x)/(2x^2+1)`

Explanation:

"Function Composition"

`(f(g(x)))'= f’(g(x))g’(x)`

`g(x)' = (2x^2+1)'= 4x`

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

`(f(g(x)))'= f’(g(x))g’(x) = 1/(2x^2+1) * 4x= (4x)/(2x^2+1)`