Finding a derivative using the FTC Part 1

[BlackMamba]

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]

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]

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

 

[HallsofIvy]

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
G(y)= \int_y^2 sin(x^2) dx[/itex]<br /> and are asked to find <br /> \frac{dG}{dy}[/itex]&lt;br /&gt; or are given&lt;br /&gt; G(x)= \int_x^2 sin(t^2)dt[/itex]&amp;lt;br /&amp;gt; and are asked to find&amp;lt;br /&amp;gt; \frac{dG}{dx}[/itex]&amp;amp;lt;br /&amp;amp;gt; then it would be a non-trivial problem (but still easy).

 

[BlackMamba]

Thanks for all of the replies, it's greatly appreciated.