# Fundamental THM of Calc?

1. Sep 8, 2009

### deteam

1. The problem statement, all variables and given/known data
Heres the question:

f(t) = Integral from c to t of a(s)ds, and h(t) = ef(t)

Compute h' (t) using:

2 fundamental theorems of Calculus

and

The Chain Rule

2. Relevant equations

Fundamental Theorems of calc

g(x) = Integral from a to x of f(t) dt

and

Integral from a to b of f(x) dx = F(b) - F(a)

and chain rule

3. The attempt at a solution

using the chain rule i know the answer is f'(t)*ef(t)

but how do you do it using the 2 fundamental thrms??

Last edited: Sep 8, 2009
2. Sep 8, 2009

### Dick

Nooo. The derivative of h(t) is f'(t)*e^(f(t)) isn't it? Now start working on the derivative of f(x).

3. Sep 8, 2009

### deteam

so basically f(t) = Integral from c to t of a(s)

f(t) = a(t) - a(c)

then f'(t) = ??

4. Sep 8, 2009

### Dick

Wrong again. f(t)=A(t)-A(c) where A(t) is an antiderivative of a(t), i.e. A'(t)=a(t). Tell me again what f'(t) is. I'll warn you in advance, I think you'll get it wrong. Prove I'm wrong about that and think about it.

5. Sep 8, 2009

### deteam

ya I understand that, part 2 of the Fund thm of calc. but can u please explain your whole process on what your doing here cause your just pulling out these rules which still don't help me understand the process of setting this up.

6. Sep 8, 2009

### Dick

I'm not 'pulling out any rules'. I'm just using the chain rule which says (h(f(x)))'=f'(x)*h'(f(x)) with h(x)=e^x (so h'(x)=e^x) and the fundamental theorem of calculus which says the integral from c to x of f(x) is F(x)-F(c) where F is an antiderivative of f. These are the rules you are told to use. Just apply those to your problem.

7. Sep 8, 2009

### deteam

alright, this isn't really helping [maybe I'm just not getting it] but thanks for trying, appreciate it.

8. Sep 8, 2009

### Dick

I appreciate the thanks, but I don't think you are trying very hard. If f(t)=A(t)-A(c) and A'(t)=a(t), what's the derivative of f(t) (remembering that the derivative of a constant is 0)? I really am trying to help.

9. Sep 8, 2009

### deteam

well since A'(t) = a(t) then wouldn't the derivative of f(t) just be A'(t) and A'(c)?

so:
f(t)=A(t)-A(c)
f'(t) = a(t) - a(c) ?

**The main problem is Im in Calc C and all this stuff was stuff I did in Calc B, which was all the way back in Feb.**

10. Sep 8, 2009

### Dick

Excuses understood. But A(c) is a CONSTANT. I don't think it's derivative will be nonzero. Hint, hint.

11. Sep 8, 2009

### deteam

ok well since its a constant then like you said the derivative is zero. [ i wasn't grasping the whole a(c) is a constant part] which leaves the derivative of f(t) = a(t) only?

and i appreciate you being patient with this, i probably would of stopped helping by now :yuck:

12. Sep 8, 2009

### Dick

Yes, absolutely.

13. Sep 8, 2009

### deteam

so now this means [ going back to the original problem] that since:
h(t) = ef(t)
and
h'(t) = f'(t)*ef(t)

h'(t) = a(t)*eA(t) -A(c)
?????

14. Sep 8, 2009

### Dick

Right. Or h'(t)=a(t)*e^(f(t)). Yes?

15. Sep 8, 2009

### deteam

yess, thanks a lot once again appreciate your patience, and 2 thumbs up from me lol