Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

"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?

share|improve this question
    
possible duplicate of When does product of derivatives equals derivative of products? –  MJD Jul 5 at 3:57
2  
@MJD This isn't a duplicate - the question here is about finding a specific $g$ to make the naive rule work given $f$. –  T. Bongers Jul 5 at 4:24
1  
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 at 5:02

2 Answers 2

up vote 9 down vote accepted

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?

share|improve this answer
    
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 at 3:45
    
Exactly right :) Good job –  Cameron Williams Jul 5 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!

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.