Question

What are the first and second derivatives of `f(x)=(ln x)^(1/2)`?

Answer 1

`f'(x) = 1/(2 x sqrt(ln x) )`

and

`f''(x) = -1/(2x^2) [1/(2sqrt(ln^3 x))+1/(sqrt(ln x))]`

Explanation:

To get the first derivative, we'll use the chain rule :

`f'(x) = (d f(x))/(dx) = d/(dx) (ln x)^(1/2) = 1/2(ln x)^(-1/2)*d/(dx) (ln x) = 1/2(ln x)^(-1/2)*1/x`

the second derivative again uses first the product rule and then the chain rule:

`f''(x) = (d f'(x))/(dx) = d/(dx) 1/2(ln x)^(-1/2)*1/x = -1/4 (ln x)^(-3/2)d/(dx) (ln x)*1/x+1/2(ln x)^(-1/2) (-1/x^2) = -1/4 (ln x)^(-3/2) 1/x^2 - 1/2 (ln x)^(-1/2) 1/x^2`

We can also choose to write these two answers in simpler forms:

`f'(x) = 1/(2 x sqrt(ln x) )`

and

`f''(x) = -1/(2x^2) [1/(2sqrt(ln^3 x))+1/(sqrt(ln x))]`