Finding a derivative using the FTC Part 1

  • Thread starter BlackMamba
  • Start date
  • #1
187
0
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:

[itex]g(x) = \int_{a}^{h(x)} f(t) dt[/itex] then, [itex]g'(x) = f(h(x)) * h'(x)[/itex]

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: [itex]G(x) = \int_{y}^{2} sin(x^2) dx[/itex]

My Solution:
[itex]G'(x) = sin(2^2) * 0[/itex]
[itex] = sin(4) * 0[/itex]
[itex]= 0[/itex]


Thanks for taking a look. :smile:
 

Answers and Replies

  • #2
AKG
Science Advisor
Homework Helper
2,565
4
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:

[tex]G(x) = \int _y ^2 \sin (x^2)\, dx[/tex]

can be rewritten:

[tex]G(x) = \int _y ^2 \sin (z^2)\, dz[/tex]

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.
 
  • #3
StatusX
Homework Helper
2,571
1
Are you sure it isn't G(y)?
 
  • #4
HallsofIvy
Science Advisor
Homework Helper
41,847
966
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]
and are asked to find
[tex]\frac{dG}{dy}[/itex]
or are given
[tex]G(x)= \int_x^2 sin(t^2)dt[/itex]
and are asked to find
[tex]\frac{dG}{dx}[/itex]
then it would be a non-trivial problem (but still easy).
 
  • #5
187
0
Thanks for all of the replies, it's greatly appreciated.
 

Related Threads on Finding a derivative using the FTC Part 1

  • Last Post
Replies
3
Views
1K
Replies
2
Views
1K
  • Last Post
Replies
12
Views
1K
  • Last Post
Replies
1
Views
748
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
13
Views
2K
Replies
5
Views
1K
  • Last Post
Replies
6
Views
2K
Replies
3
Views
2K
Top