# Differentiation of (2x)^(2x)

1. Jun 8, 2013

### FallingMan

1. The problem statement, all variables and given/known data

$$\frac{d}{dx}\, (y = (2x)^{(2x)})$$

2. Relevant equations

Chain Rule: d/dx (g * x) = g'x + x'g

3. The attempt at a solution

$$y = (2x)^{(2x)}$$

1. Take natural log of both sides.

$$ln(y) = ln(2x)·(2x)$$

2. Differentiate both sides

$$\frac{dy}{dx}\,·\frac{1}{y}\, = 2·ln(2x)+2$$

3. Substitute y in and distribute

$$\frac{dy}{dx}\, = 2x^{2x}(2·ln(2x)+2)$$

This solution is incorrect. Any help would be appreciated.

2. Jun 8, 2013

### cp255

This may just be a technicality but there are no parentheses around the (2x)^(2x) in step 3. That would be technically wrong. How do you know the solution is incorrect? Is a computer grading it or are you looking at the correct answer in a text book.

3. Jun 8, 2013

### FallingMan

Hi, cpi255. It's a computer graded problem. I made sure to add parentheses in my answer when I submitted it to their system.

I'm looking at wolframalpha's solution right now, and it seems to be different than the one I have arrived at.

4. Jun 8, 2013

### cp255

Enter this expression into Wolfram Alpha. You will see that your answer is correct.
(d/dx (2x)^(2x)) = ((2x)^(2x)(2*ln(2x) + 2)

Also are you sure in the problem the that the base was in the parentheses?

5. Jun 8, 2013

### Dick

It looks pretty ok to me. Except you meant $\frac{dy}{dx}\, = (2x)^{2x}(2·ln(2x)+2)$ in 3. right? You just missed a parentheses.

6. Jun 8, 2013

### FallingMan

Huh, I guess that's what it was! I was adding parenthesis to the top exponent and not the bottom. Damn!

Thanks a lot, cp255 and Dick!