Question

How do you find the first and second derivative of `lnx^(1/2)`?

Answer 1

`f'(x)=1/(2x)` and `f''(x)=-1/(2x^2)`

Explanation:

We use the concept of function of a function according to which,.

`d/(dx)f(g(x))=(df)/(dg)xx(dg)/(dx)`

Hence as `f(x)=lnx^(1/2)`

`f'(x)=1/x^(1/2)xx1/2x^(-1/2)=1/(2x)`

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