$\frac{d}{dx} \int_{a}^{x} f(x,t) \ dt$

Question

Does $$\frac{d}{dx} \int_{a}^{x} f(x,t) \ dt$$ equal to $f(x,x)$ by Fundamental Theorem of Calculus?

Answer 1

Negative. Consider $f(x,t) = x$ and

$$\frac{d}{dx} \int_a^x x dt = \frac{d}{dx}(x(x-a)) = 2x - a \neq x = f(x,x)$$

The problem is the integrand is also a function of $x$. You can read How do I differentiate this integral? for more.

Answer 2

that is not true. the leibniz formula is $$ \frac{d}{dx} \int_a^x f(x, t)\, dt =f(x,x) + \int_a^x \frac{\partial f(x,t)}{\partial x} \, dt.$$

Answer 3

Applying Leibniz's rule: $$ \frac{d}{dx} \int\limits_a^x f(x,t) \,dt = \int\limits_a^x f_x(x,t) \,dt + f(x,x) \frac{dx}{dx} = \int\limits_a^x f_x(x,t) \,dt + f(x,x) $$

Answer 4

I seem to remember the Fundamental Theorem of Calculus proven with regards to functions of one variable. In this case you have two, so it's definitely a good idea to be a little careful. It's always a good idea to make sure you check whether or not your problem fits into the hypotheses of the theorem and that certain things can safely be done. Another example would be interchanging the order of a limit and a function - it seems perfectly safe, but you can't do it if the function isn't continuous.

For reference, here's the Fundamental Theorem of Calculus. I don't see any provision for when your variable of differentiation is in both the integrand and the limits of integration. If I didn't know any better, I would try a few examples to see whether or not it seems to hold.

You might find it useful to keep looking for answers.... I think Leibniz' Rule holds one for you.