How do I find the derivative of this equation using the fundamental theorem of calculus? Can someone explain how to find the derivative using the fundamental theorem of calculus on the following equation?

$$g(x) = \int_{2}^{x} t^2 \sin(t) dt  $$
Here is one way to think about it. What is your usual approach to calculating a definite integral, like
$$\int_2^x t^2\sin(t)\,dt?$$

Ordinarily, you go through the following steps:

$1$. Find an antiderivative $F(t)$ of $f(t)$, where $f(t)=t^2\sin(t)$. That is, find a function $F(t)$ such that $F'(t)=f(t)=t^2\sin(t)$.

$2$. Then $\int_2^x t^2\sin(t)\,dt=F(x)-F(2)$ (you "plug in").

We are asked for the derivative of $\int_2^x t^2\sin(t)\,dt$ with respect to $x$. Since the integral is $F(x)-F(2)$, you need to differentiate $F(x)-F(2)$. The answer is $F'(x)$. But $F'(x)=f(x)=x^2\sin(x)$, and we are finished.

Note that we did not need to find an explicit formula for $F(t)$. That would be a waste of time, we would be integrating $f(t)$ to find $F(x)-F(2)$, only to turn around and differentiate it!

With a little work (integration by parts), we could in fact find an explicit antiderivative $F(t)$ of $t^2\sin(t)$. But look at the problem of finding the derivative with respect to $x$ of $\int_1^x \sin(t^2)\,dt$. You will not be able to find a "nice" function whose derivative is $\sin(t^2)$. (Neither will I.) But using the same reasoning as above, we can see that the derivative of $\int_1^x\sin(t^2)\,dt$ with respect to $x$ is $\sin(x^2)$.

Here is a fancier example. Find the derivative with respect to $x$ of $\int_1^{x^2}e^{-t^3}\,dt$. Imagine that we were able to find an explicit antiderivative $F(t)$ of $f(t)=e^{-t^3}$. Then our integral would be $F(x^2)-F(1)$. The derivative of this with respect to $x$ is $(2x)F'(x^2)$ (remember the Chain Rule). But $F'(t)=f(t)=e^{-t^3}$, and therefore our derivative is $(2x)e^{-x^6}$.  Note that at no time did we actually find an explicit antiderivative of $e^{-t^3}$.
(2017/8/6) "Amusing" silent downvotes, 5 years after this was posted. Surely for mathematical reasons... :-)


One can go back to the definition of the derivative. To wit, note that, by the first mean value theorem, $g(x+h)-g(x)=hf(z)$ with $f(z)=z^2\sin(z)$, for some $z$ depending on $x$ and $h$, such that $|z-x|\leqslant h$. Now, the function $f$ is continuous at $x$ hence, when $h\to0$, $f(z)\ldots$