# Finding a rare case where an incorrect rule works?

"A not uncommon error in calculus is to believe that the product rule for derivatives states that $(fg)' = f'g'$. If $f(x) = e^{3x}$, find a nonzero function g for which $(fg)' = f'g'$."

I believe you can find the function(s) using algebra, I got ${dy \above 1pt dx}ge^{3x} = g'*3e^{3x}$ but I don't know what to do with $g$. What would I sub in for $g$ and $g'$, or am I going about this all wrong?

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possible duplicate of When does product of derivatives equals derivative of products? –  MJD Jul 5 '14 at 3:57
@MJD This isn't a duplicate - the question here is about finding a specific $g$ to make the naive rule work given $f$. –  user61527 Jul 5 '14 at 4:24
Since the other question is about the general problem of which this is a special case, and the answers can easily be specialized to this particular case, I think it is a duplicate. The answer there specifically addresses the case where one of the two functions is given. –  MJD Jul 5 '14 at 5:02

The idea is to make use of the usual product rule (since the problem is more or less asking for when the rule $(fg)'=f'g'$ works given a certain choice of $f$): $(fg)' = f'g+fg'$. Using our definition of $f$, we see that our equation $(fg)' = f'g'$ becomes

$$3e^{3x}g+e^{3x}g' = 3e^{3x}g'$$

or equivalently (by dividing through by $e^{3x}$)

$$3g+g' = 3g' \quad \Longrightarrow\quad g' = \frac{3}{2}g.$$

Do you see how to proceed?

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Working from there I got $g = Ce^{{3 \above 1pt 2}x}$. I believe you can't solve for C here so that's my final answer. –  Cains Jul 5 '14 at 3:45
Exactly right :) Good job –  Cameron Williams Jul 5 '14 at 4:14

Try this: we want $3g'e^{3x} = 3ge^{3x} + g'e^{3x}$.

We can divide through by $e^{3x}$ to get $3g' = 3g + g'$. Now combine like terms and simplify. Here we have a differential equation, but with a little thinking, you certainly don't need a differential equations course to solve it!

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