Using the fundamental theorem of calculus find $dy/dx$

Question

A function is given by:
$$y(x) = \int_0^{x^2}t\sin(t)dt$$ Find $\frac{dy}{dx}$ using the fundamental theorem of calculus.

This task really confused me, because - in the theorem - the integral is given in terms of $x$, not in terms of $x^2$. Therefore, I don't know how to proceed. One way to solve this I could think of is to change the integration bounds - to have only an $x$ up there, but I can' think of any way to do this.
Would you mind giving me a hint on solving this? I have a hunch that it may have something to do with the chain rule, but still - I have never seen a problem like this.

Answer 1

Hint. The chain rule gives $$ (F(x^2))'=2x \cdot F'(x^2)=2x \cdot f(x^2) $$ where $$ F'(t)=f(t). $$ Can you see how to apply this?

Answer 2

Well, in general:

$$\frac{\partial}{\partial x}\left(\int_\text{y}^{x^\text{n}}\text{f}\left(t\right)\space\text{d}t\right)=\text{n}\cdot x^{\text{n}-1}\cdot\text{f}\left(x^\text{n}\right)\tag1$$

Answer 3

your integral is given by $$-x^2\cos(x^2)+\sin(x^2)$$ and the first derivative is $$2x^3\sin(x^2)$$