# Fundamental THM of Calc?

deteam

## Homework Statement

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

## Homework 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

## 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??

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

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

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

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

then f'(t) = ??

Homework Helper
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.

deteam
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.

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.

Homework Helper
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.

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.

deteam
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.

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

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

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.

deteam
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.

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 I am in Calc C and all this stuff was stuff I did in Calc B, which was all the way back in Feb.**

Homework Helper
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 I am in Calc C and all this stuff was stuff I did in Calc B, which was all the way back in Feb.**

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

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

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:

Homework Helper
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?

Yes, absolutely.

deteam
Yes, absolutely.

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)
?

Homework Helper
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)
?

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

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

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