# Find dy/dx of y= the square root of ln x

1. Sep 29, 2012

### Interception

1. The problem statement, all variables and given/known data
Find dy/dx when y=$\sqrt{ln x}$

2. Relevant equations
d/dx of ln x is equal to 1/x times d/dx of x.

3. The attempt at a solution
I tried to raise the ln x to the 1/2 power instead of keeping it under a square root sign, but I had no luck. I'm struggling with Calculus. I would very much appreciate some help.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Sep 29, 2012

### sharks

To visualize the problem in an easier way, let u = ln x. Then proceed. First, differentiate the power, then the term/s under the square root.

3. Sep 29, 2012

### jmcelve

This strategy for solving the problem won't work because (ln(x))^a ≠ln(x^a) = a*ln(x). As suggested by sharks, u substitution is the best strategy here.

4. Sep 30, 2012

### danielu13

Using u-sub:

$\frac{dy}{dx}$$\sqrt{lnx}$ | u=lnx

=$\sqrt{u}$'*u'

=$\frac{1}{2\sqrt{lnx}}$*$\frac{1}{x}$

=$\frac{1}{2x\sqrt{lnx}}$

It's been a while since I've done derivatives, but I think this should be the correct way of working it.

5. Sep 30, 2012

### dextercioby

There's a nice trick at hand.

$$y^2 (x) = \ln x$$

Differentiate both sides with respect to x.