$$f(x) = \frac{\ln(x)}{\sqrt x}$$
I am trying to take the derivative of this.
What I have is:
$$f'(x)=\frac {\frac1x \sqrt x - \frac 1{2\sqrt x} \ln(x)}{x}$$
simplifying, I get:
$$-0.5\cdot\frac {\ln(x)}{x^{3/2}} + x^\frac {-1}{2}$$
-- At this point my answer differs from the solution for the term on the right hand side... I get $x^{-1/2}$ and they have $x^{-3/2}$. But I really don't understand how they got that.
And then, if it is correct up to here, I'm not sure how to solve it. If anyone could help me I would really appreciate it. Thanks!
We obtain
$$f'(x)=\frac {\frac1x \sqrt x - \frac 1{2\sqrt x} \ln(x)}{x}=\frac{\sqrt x}{x^2}-\frac 1{2x\sqrt x} \ln(x)=\frac{1}{x\sqrt x}-\frac 1{2x\sqrt x} \ln(x)$$
which agrees with the solution you are looking for.