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. 

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.
