Gradient of f: R^2 -> R Defined by Integral Equation

[johnson12]

Define $$f: R^{2} \rightarrow R , by f(x,y) = \int^{sin(x sin(y sin z))}_{a} g(s) ds$$


where g:R -> R is continuous. Find the gradient of f.


I tried using the FTC, and differentiating under the integral, but did not get anywhere,

thanks for any suggestions.

 

[HallsofIvy]

Yes, the FTC, together with the chain rule should work. Basically, you are saying that
$$f(x,y)= \int_0^u(x,y) g(s)ds$$
where u(x,y)= x sin(x sin(y sin(x))).
$$\frac{df}{du}= g(u)$$
and
$$\frac{\partial f}{\partial x}= g(u)\frac{\partial u}{\partial x}$$
$$\frac{\partial f}{\partial y}= g(u)\frac{\partial u}{\partial y}$$

So the question is really just: What are $\partial u/\partial x$ and $\partial u/\partial y$f?