Easy Derivative - Not easy for me, though

[fiziksfun]

1. Derivate x^(2sinx)


3. I used implicit differentiation because there's an x in the base and in the exponent.
I have all of my work in the picture (if you can't see it in the attachment):

http://i12.tinypic.com/7x1sn44.jpg


I don't get the same answer as the correct answer which is 2cosx(x^2sinx), and I can't figure out why!

Am I doing something wrong, or is it possible the answer is wrong !?

 

[Moridin]

Try substituting t = 2 sin x and you will get x^t where t(x). Differentiate x^t. What are you going to do if you get any t':s in your equation?

 

[rocomath]

i'm getting the same answer as you

 

[fiziksfun]

ya i just i checked my answer using a graph - its right and the other answer is wrong - thanks!

 

[rocomath]

anytime :-]

 

[dynamicsolo]

ya i just i checked my answer using a graph - its right and the other answer is wrong - thanks!

Whoever provided the 'correct' answer just treated it like d/dx (a^u), which only works for a constant a > 0. When the base of the exponential function is itself a variable or function, logarithmic differentiation is called for, which is what you did. (I sure hope that wasn't a TA or instructor who wrote that 'answer'...)