Finding the Derivative of (2x)^(2x) using the Chain Rule

[FallingMan]

Homework Statement

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

Homework Equations

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

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.

 

[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 textbook.

 

[FallingMan]

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 textbook.

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.

 

[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?

 

[Dick]

Homework Statement

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

Homework Equations

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

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.

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.

 

[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!