Fundamental Theorems of Calculus and Chain Rule: Computing h'(t)

In summary: sorry for being so dense.and i appreciate you being patient with this, i probably would of stopped helping by now... sorry for being so dense.
  • #1
deteam
24
0

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 Ruleany 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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  • #2
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
Dick said:
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
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
Dick said:
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
deteam said:
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
Dick said:
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
deteam said:
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
Dick said:
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
deteam said:
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
Dick said:
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
deteam said:
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
Dick said:
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
deteam said:
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
Dick said:
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
 

1. What is the Fundamental Theorem of Calculus?

The Fundamental Theorem of Calculus is a mathematical theorem that connects the concepts of differentiation and integration. It states that if a function is continuous on a closed interval, then its definite integral on that interval can be evaluated using an antiderivative of the function.

2. Why is the Fundamental Theorem of Calculus important?

The Fundamental Theorem of Calculus is important because it provides a fundamental tool for solving problems in calculus. It allows us to easily determine the area under a curve and calculate other important values, such as average values and total change.

3. How is the Fundamental Theorem of Calculus applied in real life?

The Fundamental Theorem of Calculus has many real-life applications, such as in physics, engineering, and economics. It is used to calculate areas, volumes, and rates of change in real-world situations.

4. What are the two parts of the Fundamental Theorem of Calculus?

The Fundamental Theorem of Calculus has two parts: the first part, also known as the Fundamental Theorem of Calculus Part I, deals with finding the antiderivative of a function, while the second part, or the Fundamental Theorem of Calculus Part II, involves evaluating definite integrals.

5. Can the Fundamental Theorem of Calculus be extended to multivariable calculus?

Yes, the Fundamental Theorem of Calculus can be extended to multivariable calculus. In this case, it is known as the Multivariable Fundamental Theorem of Calculus, and it involves integrating over a region in space instead of an interval on a number line.

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