xe^(sin(x)

http://tinyurl.com/7dvcbtb

I looked at the steps but I don't understand how that chain rule worked.

Shouldn't f(u) = usin (x) ?

Answers
I need to differentiate it.
ecofox
14 March 2012
Hi ecofox, There are two rules you need to differentiate this, the product rule: d (u*v)/dx=udv/dx+vdu/dx. And the chain rule: dy/dx=dy/du * du/dx. The second is a tricky one to understand but imagine dy/du and du/dx as fractions and see how they'd cancel out when you multiply them. You should use th chain rule on e^sin(x), i.e y=e^u and u=sin(x). You almost had it above but you need to choose f(u) and u into two function you know how to differentiate, i.e f(u)=e^u and u=sin(x). I hope that all makes sense, if not I'm happy to try and explain it more clearly.
sinsua
14 March 2012
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