How to apply the fundamental theorem to partial derivatives?

[Eclair_de_XII]

Homework Statement

"Under mild continuity restrictions, it is true that if $F(x)=\int_a^b g(t,x)dt$,

then $F'(x)=\int_a^b g_x(t,x)dt$.

Using this fact and the Chain Rule, we can find the derivative of

$F(x)=\int_{a}^{f(x)} g(t,x)dt$

by letting

$G(u,x)=\int_a^u g(t,x)dt$,

where $u=f(x)$."

Given:

$F(x)=\int_{0}^{x^2} \sqrt{t^4+x^3}dt$,

find $\frac{dF}{dx}$.

Homework Equations

$\frac{dF}{dx}=\frac{dF}{dt}\frac{dt}{dx}+\frac{dF}{du}\frac{du}{dx}$

The Attempt at a Solution

$\frac{dF}{du}=\frac{d}{du} \int_{0}^{x^2}\sqrt{t^4+x^3}dt=\sqrt{(x^2)^4+x^3}=\sqrt{x^8+x^3}$
$\frac{du}{dx}=\frac{d}{dx}(x^2)=2x$ by letting $u=x^2$

$\frac{dF}{dt}=F'(t)=\int_{0}^{x^2} g_x(t^4+x^3)dt=\int_{0}^{x^2} \frac{3x^2}{2\sqrt{t^4+x^3}}$

I know that: $\frac{dF}{dx}=2x\sqrt{x^8+x^3}+\int_{0}^{x^2} \frac{3x^2}{2\sqrt{t^4+x^3}}\frac{dt}{dx}$

But, I'm at a loss at how to calculate: $\frac{dt}{dx}$, because I don't know how to define $t$, so I don't know how to differentiate it with respect to $x$.

 

[Delta2]

in short the way you doing it is t=x but you can't replace that inside the integral. In short the result is
$$2x\sqrt{x^8+x^3}+\int_{0}^{x^2}\frac{3x^2}{2\sqrt{t^4+x3}}dt$$.

However in order to be more accurate, we have to define $F(x)=G(u(x),x)$ with $G(u,v)=\int_0^{u}\sqrt{t^4+v^3}dt$ therefore

$$\frac{dF}{dx}=\frac{\partial G}{\partial v}\frac{\partial v}{\partial x}+\frac{\partial G}{\partial u}\frac{\partial u}{\partial x}$$.

with $u(x)=x^2$ $v(x)=x$.

Calculating the above expression you ll get the result.

 

[Eclair_de_XII]

$v(x)=x$

So $\frac{∂G}{∂v}\frac{∂v}{dx}$ corresponds to $\int_{0}^{x^2} \frac{3x^2}{2\sqrt{t^4+x^3}}$ and $\frac{∂G}{∂u}\frac{∂u}{∂x}$ corresponds to $2x\sqrt{x^8+x^3}$? And $v'(x)=1$, so it's just $\int_{0}^{x^2} \frac{3x^2}{2\sqrt{t^4+x^3}}+2x\sqrt{x^8+x^3}$. I guess that makes sense; I mean, I just need to perform the proper substitutions and such. Anyway, thank you for your help, Delta.