Erroneous application of the quotient rule

[greenpick]

Hello everyone, first time poster here. I've been a lurker for about a week, but finally decided to join because I cannot for the life of me figure out this problem.

Homework Statement

Zach has trouble with the Quotient Rule; he thinks that d/dx (f(x)/g(x)) =
f′(x)/g′(x). On his last calculus test, Zach applied this erroneous rule to a quotient in which g(x) = x.
Somehow, he managed to get the right answer. What are all the possibilities for the function f(x)?

Homework Equations

d/dx (f(x)/g(x)) = f′(x)/g′(x) (Zach's rule)
d/dx (f(x)/g(x)) = [f′(x)g(x) - f(x)g′(x)] / (g(x))^2 (actual quotient rule)

The Attempt at a Solution

I first attempted to set the derivative using Zach's equation equal to the derivative using the quotient rule.
After simplification, I got
f(x) = (x-x^2)f′(x)
I am not really sure what to do at this point however. I am assuming it is something either really obvious, or really not-so-obvious. I tried using a substitution, and even an hour of trial and error, but I am just not seeing it. Am I going to have to use the limit definition of the derivative? Any help would be greatly appreciated.

 

[n!kofeyn]

Have you learned integration yet? This amounts to being a differential equation. What I mean is that you can solve for f by performing the following integration:
[tex]\int \frac{f'(x)}{f(x)} \,dx = \int \frac{1}{x(1-x)} \,dx[/tex]

 

[Dick]

f(x) = (x-x^2)f′(x) looks correct. No, you don't need any limits. Now it's a separable differential equation. Do you know anything about solving differential equations?

 

[n!kofeyn]

f(x) = (x-x^2)f′(x) looks correct. No, you don't need any limits. Now it's a separable differential equation. Do you know anything about solving differential equations?

Isn't this exactly what I just posted?

 

[Dick]

Isn't this exactly what I just posted?

Of course it is. I didn't see your post before I posted. Want me to delete it?

 

[n!kofeyn]

Of course it is. I didn't see your post before I posted. Want me to delete it?

Haha. Sorry. No problem. There's no reason to delete. :)

 

[greenpick]

Wow, thank you guys for the quick replies.

Fortunatly, we just started learning antiderivatives and the basics of integrals in class the end of last week. UNfortunatly, we haven't gone over differential equations with integrals. It appears that I need to find the antiderivative of 1 / [x(1-x)]. Is this correct? It looks like I am going to have to make a substitution.
This is just a shot in the dark, but would I...
let u = x(1-x) so that [tex]\int\frac{1}{x(1-x)}dx[/tex] becomes [tex]\int\frac{1}{u}dx[/tex] ?
But would I need to change dx to du in the second integral? would it then become
[tex]\int\frac{1}{u}du[/tex] ?
Am I on the right track?

 

[n!kofeyn]

First notice that
[tex]\frac{1}{x(1-x)} = \frac{1}{x} + \frac{1}{1-x}[/tex]
Then
[tex]\begin{align*}
\int \frac{f'(x)}{f(x)} \,dx &= \int \frac{1}{x(1-x)} \,dx \\
\ln[f(x)] &= \int \frac{1}{x} \,dx + \int \frac{1}{1-x} \,dx
\end{align*}[/tex]

Can you finish it now?

 

[greenpick]

Ah, I did not even notice that log derivative.
So I would have
[tex]\int\frac{1}{x}dx[/tex] + [tex]\int\frac{1}{1-x}dx[/tex] = ln(abs(x)) - ln(abs(1-x))
ln(abs(x)) - ln(abs(1-x)) = ln(f(x)) = ln(abs(x/(1-x)))
f(x) = e[tex]^{ln(abs(x/(1-x)))}[/tex] = abs(x/(1-x)) + C ?

 

[n!kofeyn]

Ah, I did not even notice that log derivative.
So I would have
[tex]\int\frac{1}{x}dx[/tex] + [tex]\int\frac{1}{1-x}dx[/tex] = ln(abs(x)) - ln(abs(1-x))
ln(abs(x)) - ln(abs(1-x)) = ln(f(x)) = ln(abs(x/(1-x)))
f(x) = e[tex]^{ln(abs(x/(1-x)))}[/tex] = abs(x/(1-x)) + C ?

Yep!

 

[greenpick]

Wow! Thank you guys so much that helped.

 

[n!kofeyn]

No problem! Congratulations on solving your first differential equation.