Fundamental Theorem of Calculus with multiple variables I'm stuck on this multivariable equation:
$$
\frac{d}{dx}\left(\int^x_af(g(b,t),t)dt\right)
$$
where a and b are just constants.
If this involved a single variable, it looks like one would just apply the fundamental theorem of calculus. Is there an equivalent for multiple variables.
I know that the answer should just be
$$
f(g(b,x),x)
$$
but I'm hoping someone can explain / walk me through. Is there maybe some rule that lets me pass the $\frac{d}{dx}$ into the integral?
Thanks
 A: The function under the integral can be seen as a function of one variable $t$, and then the fundamental theorem applies directly. That is, if we define $h(t) = f(g(b,t),t)$, then $$\frac{d}{dx}\int_a^x h(t) \, dt = h(x),$$ and if we unwind the definition this is says that $$\frac{d}{dx}\int_a^x f(g(b,t),t) \, dt = f(g(b,x),x).$$
A: This doesn't actually need multiple variables, and could be deduced from the single variable FTC. You also don't want to pass the limit inside the integral since the limit of integration depends on $x$ (which you're trying to take the limit of).
Recall that if you have a single variable function $h(t)$ then:
$$\dfrac{d}{dx} \int_a^x h(t)dt=h(x) $$
in your case, for fixed $b$, take $h(t)=f(g(b,t),t)$. Notice this is just a single variable function. The fact that it is actually a composition of two single variable functions and that there's an extra constant $b$ doesn't change the fact that it's still a single variable function and hence the above analysis still applies:
$$\dfrac{d}{dx} \int_a^x h(t)dt=h(x) \implies \dfrac{d}{dx} \int_a^x f(g(b,t),t)dt=f(g(b,x),x)$$
