Derivative of x^(x^2) with respect to x

[nobahar]

Homework Statement

Differentiate $$x^{x^2}$$, with respect to x

Homework Equations

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

The Attempt at a Solution

I arrived at... (ready?)

(ta dah!): $$x^{x^2}.(2x.\ln(x)+x)$$

I'm pretty confident this is wrong...

I went $$y(x)=x^{x^2}$$, then took the natural logarithm of both sides

Since $$\ln(x^{x^2}) = x^2.\ln(x) and \frac{d}{d(y(x))} (ln (y(x))) = \frac{1}{y(x)}$$

I got:

$$\frac{1}{y(x)} . \frac{d(y(x))}{dx} = \frac{d}{dx} (x^2.\ln(x))$$

$$\frac{1}{y(x)} . \frac{d(y(x))}{dx} = 2x.\ln(x)+\frac{x^2}{x}$$

$$\frac{d(y(x))}{dx} = 2x.\ln(x)+\frac{x^2}{x} . x^{x^2} = x^{x^2}.(2x.\ln(x)+x)$$

As I said, I think this is wrong. I've been working through examples all day and figured I might be able to come back to it, and hopefully figure it out (that is my excuse, and I'm sticking with it! ); but since I don't have easy access to a computer, I thought I might ask you guys now, and check back tomorrow to see if anyone has offered any help.
Thanks everyone.

 

[snipez90]

Eh I think that is right, though I did it in my head. It's easy if you write it as the exponential of (x^2)(ln x), from which you get the original expression times the derivative of (x^2)(ln x) which by the product rule is 2x(ln x) + x.

 

[Cyosis]

I don't see why you think this would be wrong, because it's not.

 

[karkas]

Yup, I think it's pretty correct.

 

[nobahar]

Snipez90, Cyosis & Karkas: Thankyou.
The question was from a multiple choice "quiz" that I saw, and I just didn't recognise the answer that I got (above) from being one of the possibilities: that's why I thought it was wrong. But it seems its not! Thanks for the responses everyone!