Can you apply the fundamental theorem of calculus with the variable inside the integrand? I was wondering if I can apply the FTOC: $\frac { d }{ dx } \left( \int _{ a }^{ x }{ f(t) } dt \right) =f(x)$ to an implicit function with the variable being differentiated implicit in the function. 
For example: $\frac { d }{ dx } \left( \int _{ a }^{ x }{ xf(t) } dt \right)$ or any function in the form: $\frac { d }{ dx } \left( \int _{ a }^{ x }{ f(x,t)f(t) } dt \right) $ and if so, what would we get. 
 A: By definition
$$\frac{d}{dx}\int_a^xf(x,t)dt=\lim_{\Delta x\to 0}\frac{1}{\Delta x}\left[\int_a^{x+\Delta x}f(x+\Delta x,t)dt-\int_a^x f(x,t)dt\right]$$
$$=\lim_{\Delta x\to 0}\frac{1}{\Delta x}\left[\int_a^{x+\Delta x}\left(f(x,t)+\frac{\partial f}{\partial x}\Delta x\right)dt-\int_a^x f(x,t)dt\right]$$
$$=\lim_{\Delta x\to 0}\frac{1}{\Delta x}\left[\int_a^{x+\Delta x}f(x,t)dt-\int_a^x f(x,t)dt + \int_a^{x+\Delta x}\frac{\partial f}{\partial x}\Delta x dt\right]$$
$$=\lim_{\Delta x\to 0}\frac{1}{\Delta x}\left[\int_x^{x+\Delta x}f(x,t)dt + \int_a^{x+\Delta x}\frac{\partial f}{\partial x}\Delta x dt\right]$$
$$=\lim_{\Delta x\to 0}\frac{1}{\Delta x}\left[f(x,x)\Delta x + \Delta x \int_a^{x+\Delta x}\frac{\partial f}{\partial x}dt\right]$$
$$=f(x,x)+\lim_{\Delta x\to 0} \int_a^{x+\Delta x}\frac{\partial f}{\partial x}dt$$
$$=f(x,x)+\int_a^x \frac{\partial}{\partial x}f(x,t)dt$$
In conclusion, you will only get the first term if you apply the FTOC blindly. Because the integrand contains $x$, you will have the second extra term.
A: In mathematics any question of the form:

Can we do X?

always has the answer:

If you have proven that you can.

In particular if you ask:

Can you apply theorem X to Y?

the answer is just:

As long as Y satisfies the conditions required by theorem X.

In your case, the fundamental theorem of calculus does not apply when the integrand involves the limit of integration.
You may be wondering whether the theorem can somehow be extended to cover that case. The simple answer is that it cannot, and nothing more can be said unless we know exactly what kind of integrand it is.
