Differentiation problem

[dbzgtjh]

can anyone tell me how to differentiate this function f(x)=2sin(x)+2x^x?

[matt grime]

x^x is the same as exp{xlogx}

you can now differentiate using the basic results you've been taught.

[kastarov]

Hello, dbzgtjh!
First of all, we must know how to differentiate this function:
[f(x)]^g(x) = h(x), so, ln{[f(x)]^g(x)} = ln h(x), is equal to
g(x)lnf(x) = ln h(x)
differentiating implicitly, we have, g'(x)lnf(x) + g(x)f'(x)/f(x) = h'(x)/h(x),
therefore,
h'(x) = h(x) [ g'(x)lnf(x) + g(x)f'(x)/f(x) ]
In this case g(x) = x = f(x), so
h(x) = x^x, so h'(x) = (x^x)(lnx + 1)
Therefore,
the derivative of function f is f'(x) = 2cosx + 2(x^x)(lnx + 1).

[arildno]

Hello, dbzgtjh!
Differential function is f'(x) = 2cos(x)+2e^(xlnx). Any doubt look for a textbook, for example:
Calculus; Stewart, James. v.1
This is incorrect; you must apply the chain rule on the last term.