Fundamental THM of Calc?

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  • #1
deteam
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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


any help please ?



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??

Help Please
 
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Answers and Replies

  • #2
Dick
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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
deteam
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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) = ??
 
  • #4
Dick
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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
deteam
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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.
 
  • #6
Dick
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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.
 
  • #7
deteam
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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.
 
  • #8
Dick
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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.
 
  • #9
deteam
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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.**
 
  • #10
Dick
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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.
 
  • #11
deteam
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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:
 
  • #12
Dick
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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.
 
  • #13
deteam
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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)
?
 
  • #14
Dick
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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?
 
  • #15
deteam
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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
 

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