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#1 2005-12-02 13:56:26
- mikau
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tough problem
I just can't get this one to work.
Let f(x) = e^bx and g(x) = e^ax. Find the value of b in terms of a such that [f(x)/g(x)]' = f'(x)/g'(x)
f'(x) = e^bx * (b+x) g'(x) = e^ax * (a+x) Correct? Did I do that right?
So f'(x)/g'(x) = e^(bx - ax) * (b+x)/(a+x) Correct again?
To find the deriviave of f(x)/g(x) I'll divide them first. e^bx/e^ax = e^(bx - ax) Right?
Now we'll find the derivitive:
e^(bx - ax) * ( b + x - ( a + x) ) which simplifies to e^(bx-ax) * (b - a) Correct?
So we found the quotient of the derivitives, and the derivative of the quotient. (thats fun to say! :-D ) If they are equal:
e^(bx - ax) * (b+x)/(a+x) = e^(bx-ax) * (b - a)
Fortunatly the e^(bx - ax) appears on both sides. So we can divide both sides by e^(bx -ax) to get:
(b+x)/(a+x) = b - a
We can now solve for b in terms of a and x, but the book has the answer only in terms of the variable x. If I could replace x with an expression involving a that would do it but I don't know where to find one. :-(
A logarithm is just a misspelled algorithm.
#2 2005-12-02 14:51:20
- ryos
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Re: tough problem
f'(x) = e^bx * (b+x) g'(x) = e^ax * (a+x) Correct? Did I do that right?
Nope. (e^bx)′ = (e^u)′(bx)′ = be^bx. (That's the chain rule.) The same goes for e^ax.
We know that (e^bx / e^ax)′ = be^bx / ae^ax. That is,
(b-a)e^(bx-ax) = be^bx / ae^ax
At this point, I got lazy and told my calculator to solve that for b. It spat this out:
b = a^2 / (a - e^2ax)
That's as far as I can go with it.
El que pega primero pega dos veces.
#3 2005-12-02 15:34:52
- mikau
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Re: tough problem
Hmm... I'll have to review the chain rule. But isn't that what I did? The derivative of bx is b * the derivitive of x (which is 1) + x * the derivitive of b (which is also 1) so it should be e^u (x + b) right?
Thats not the answer. My book wants the answer only in terms of a and constants. You can't use x. The answer to the problem is: a^2 / (a-1)
Wierd huh?
A logarithm is just a misspelled algorithm.
#5 2005-12-02 16:21:29
- mikau
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Re: tough problem
Well I guess we must assume a and b are constants. I don't think we can differentiate with more then two variables.
So I guess it would have to be with respect to the independant variable. But if they are constants then their derivatives are zero. That makes a differance. I just hope its not a false assumption.
in this case if we have to find the derivative of u = bx we have to multiply b by the derivitive of x which is 1. And x * the derivitive of b. But if b is a constant then its slope is 0.
Ok I'll try messing with that to see if I can rearrange it into the desired form. But how am I supposed to know b and a were constants?
A logarithm is just a misspelled algorithm.
#6 2005-12-02 17:20:14
- ryos
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Re: tough problem
But how am I supposed to know b and a were constants?
Mathematical convention. In the expression ax, a is the coefficient of x.
BTW, since a is not a function (but a constant), you use the constant multiple rule. That is, (a*f(x))′ = a * f′(x).
Hey, I've got it! That was a tricky bit of algebra.
We have (b-a)e^(bx-ax) = be^bx / ae^ax
Now, (b-a)e^(bx-ax) = (e^bx / e^ax) * (b-a)
and be^bx / ae^ax = (b/a) * (e^bx / e^ax)
Our expression is now (e^bx / e^ax) * (b-a) = (b/a) * (e^bx / e^ax)
Cancel out the e's to get b-a = b/a. Solve for b:
b = ab - a²
a² = ab - b
a² / (ab - b) = 1
a²b / (ab - b) = b
b = a² / (a-1)
The stuff you come up with, I swear...
El que pega primero pega dos veces.
#7 2005-12-02 17:32:42
- mikau
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Re: tough problem
The stuff you come up with, I swear...
The stuff I come up with? Or are you talking to yourself like all mathematicians do! Anyways I didn't invent that problem if thats what your thinking.
Last edited by mikau (2005-12-02 17:33:08)
A logarithm is just a misspelled algorithm.
#8 2005-12-02 17:44:30
- mikau
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Re: tough problem
Aha! I just did the problem and it worked out fine. Yeah the derivitive of the confusion ;-) was simply that I thought a and b were variables and not constants. So I gave them a derivative of 1 instead of zero. Thanks for your help ryos! You RYOCK! hehehe! :-P
You know when you get stuck on something like this for an ENTIRE DAY, its a pain in the neck but once you find out where your error was, you never forget the mistake you made and rarely make it again.
A logarithm is just a misspelled algorithm.
