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 yf?