I want to evaluate
$$\frac{d}{dx}\int_0^{x^2}\sin(t^2)dt$$
by the first part of the fundamental theorem of calculus. I see two different ways of evaluating this problem.
Method 1:
Since the first part of the fundamental theorem of calculus states that
$$\frac{d}{dx}\int_a^{x}f(t)dt=f(x)$$
we have that
$$\frac{d}{dx}\int_0^{x^2}\sin(t^2)dt=2x\sin(x^2)\tag{1}$$
Method 2:
The above analysis is wrong since you need to substitute $x^2$ for $t^2$. The correct answer is
$$\frac{d}{dx}\int_0^{x^2}\sin(t^2)dt=2x\sin((x^2)^2)=2x\sin(x^4)\tag{2}$$
Is $(2)$ the correct solution to this problem? What is the best way of explaining why $(2)$ is correct?
Best Answer
Name
$$G(x) = \int_0^{x}\sin(t^2)dt$$ and $F(x) = x^2$
You have
$$H(x) = \int_0^{x^2}\sin(t^2)dt = (G \circ F)(x).$$
Apply the chain rule. You're looking to compute $$H^\prime(x) = F^\prime(x)(G^\prime \circ F)(x).$$
As you noticed $$G^\prime(x) = \sin(x^2).$$
Hence $$H^\prime(x) = 2x \sin(x^4)$$
A general advice: give names to the objects (maps here) and use theorems based on those names. This ease the understanding when you're starting to use theorems new to you.