Derivative of f(X)^g(X) where both base and exponent are functions

[verdverm]

I am having trouble finding the rule for the (partial) derivative of an expression like

y = f(X)^{g(X)}

can anyone help?

 

[micromass]

Hint:

$$f(x)^{g(x)}=e^{f(x)g(x)}$$

 

[Mark44]

Make that e^{g(x) * ln(f(x))} and I'll agree.

 

[micromass]

Make that e^{g(x) * ln(f(x))} and I'll agree.

Oh my, I should pay better attention while posting

 

[verdverm]

hmm

I was just looking at:
wolfram alpha for
$$d/dx \ f(x,y)^{g(x,y)} = f^{(1,0)}(x,y)g(x,y)f(x,y)^{g(x,y)-1} + g^{(1,0)}(x,y)f(x,y)^{g(x,y)}log(f(x,y) \\<br /> d/dy \ f(x,y)^{g(x,y) } = f^{(0,1)}(x,y)g(x,y)f(x,y)^{g(x,y)-1} + g^{(0,1)}(x,y)f(x,y)^{g(x,y)}log(f(x,y) \\<br /> d/dx \ e^{f(x,y)g(x,y)} = f^{(1,0)}(x,y)g(x,y)e^{f(x,y)g(x,y)} + f(x,y)g^{(1,0)}(x,y)e^{f(x,y)g(x,y)}$$

and was unsure what the $$d/dx \ f(x,y) \ is\ f^{(1,0)or(0,1)}(x,y)$$ meant in the previous equations.
Is it just the partial derivative? or partial at the point (1,0) or (0,1)?

 

[verdverm]

my intuition is that it's a bit vector to show which variable the derivative is with respect to