# Possible mistake during differentiation? Please check

## Main Question or Discussion Point

For a function f(x), I have to determine intervals of increase/decrease, find local max(s)/min(s), and find intervals of concavity. The first thing I'm doing in this is to write out f'(x) and f''(x).

f(x) = $ln(x)/\sqrt{x}$

For f'(x), I used the quotient rule and received f'(x) = (($\frac{1}{x}$$\sqrt{x}$)-($\frac{-\sqrt{x}}{2}$ln(x))) / 2

However, I plugged f(x) into wolfram alpha and it gave me: $\frac{2-ln(x)}{2x^{3/2}}$

I don't understand the difference? I thought I had done this correctly but apparently not? Wolfram alpha used the product rule. Is there some kind of algebraic gymnastics I'm forgetting about? I really want to understand where my error was made, not just which is the correct answer. Thanks!

In using the quotient rule you are not differentiating $\sqrt{x}$ corretly and the quotient rule requires you to divide by the square of the denominator.

gb7nash
Homework Helper
f(x) = $ln(x)/\sqrt{x}$

For f'(x), I used the quotient rule and received f'(x) = (($\frac{1}{x}$$\sqrt{x}$)-($\frac{-\sqrt{x}}{2}$ln(x))) / 2
I see a couple of issues here:

1) How did you get $\frac{-\sqrt{x}}{2}$ in the numerator? What is the derivative of $\sqrt{x}$ ?

2) How did you get 2 in the denominator? $(\sqrt{x})^{2} =$ ?

Using quotient rule, you should get $\frac{\frac{\sqrt{x}}{x} - \frac{\ln{x}}{2 \sqrt{x}}}{x}$ which simplifies to what you got from WA. It looks like you messed up on the derivative of $\sqrt{x}$ and on the bottom of the quotient rule. http://en.wikipedia.org/wiki/Quotient_rule

I see a couple of issues here:

1) How did you get $\frac{-\sqrt{x}}{2}$ in the numerator? What is the derivative of $\sqrt{x}$ ?

2) How did you get 2 in the denominator? $(\sqrt{x})^{2} =$ ?
1.
1. $\sqrt{x}$ = x$^{1/2}$
2. Using the power rule I bring the 1/2 out as a coefficient, and subtract one from the numerator : $\frac{1}{2}$$x^{-1/2}$
3. I simplified to : $\frac{-\sqrt{x}}{2}$