# Finding a derivative using the FTC Part 1

I have a problem which asks me to find the derivative using Part 1 of the Fundamental Theorem of Calculus.

So I know that the FTC says that if:

$g(x) = \int_{a}^{h(x)} f(t) dt$ then, $g'(x) = f(h(x)) * h'(x)$

I've got what appears to be an easy problem, maybe too easy and because of that I think I'm doing something wrong. Below is the problem and my solution. If someone could just varify if I did it correctly or not, I would greatly appreciate it.

PROBLEM: $G(x) = \int_{y}^{2} sin(x^2) dx$

My Solution:
$G'(x) = sin(2^2) * 0$
$= sin(4) * 0$
$= 0$

Thanks for taking a look.

AKG
Homework Helper
Yes, it's right. Note that G(x) basically depends only on y, so G is a constant w.r.t. x which is why it's no surprise that it's derivative w.r.t. x is 0. It's even clearer if you recognize that the x's in the integrand are just dummy variables so:

$$G(x) = \int _y ^2 \sin (x^2)\, dx$$

can be rewritten:

$$G(x) = \int _y ^2 \sin (z^2)\, dz$$

Also, the middle part of: "sin(22)*0 = sin(4)*0 = 0" is unnecessary since regardless of the fact that 22 = 4, the whole thing is something times 0, so it's just 0.

StatusX
Homework Helper
Are you sure it isn't G(y)?

HallsofIvy
Homework Helper
I'm with StatusX! Any integral of "f(x)dx" with numerical limits is just a constant. The derivative of any constant function is, of course, 0.

If you are given
[tex]G(y)= \int_y^2 sin(x^2) dx[/itex]