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#1 2005-12-01 14:56:26

mikau
Member
Registered: 2005-08-22
Posts: 1,504

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.

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#2 2005-12-01 15:51:20

ryos
Member
Registered: 2005-08-04
Posts: 394

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.

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#3 2005-12-01 16:34:52

mikau
Member
Registered: 2005-08-22
Posts: 1,504

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.

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#4 2005-12-01 16:58:06

mikau
Member
Registered: 2005-08-22
Posts: 1,504

Re: tough problem

And btw something thats also confusing me. If we differentiate:

u = bx

we get

du = b + x right? but does this mean b+ x is dx, bx? Both? Neither? I'm confused. x_x

Last edited by mikau (2005-12-01 16:59:00)


A logarithm is just a misspelled algorithm.

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#5 2005-12-01 17:21:29

mikau
Member
Registered: 2005-08-22
Posts: 1,504

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.

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#6 2005-12-01 18:20:14

ryos
Member
Registered: 2005-08-04
Posts: 394

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...:P


El que pega primero pega dos veces.

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#7 2005-12-01 18:32:42

mikau
Member
Registered: 2005-08-22
Posts: 1,504

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 (2005-12-01 18:33:08)


A logarithm is just a misspelled algorithm.

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#8 2005-12-01 18:44:30

mikau
Member
Registered: 2005-08-22
Posts: 1,504

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.

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#9 2005-12-02 04:59:10

gengen
Member
Registered: 2005-11-29
Posts: 1

Re: tough problem

help me plz
lol

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