What Was the Function g in This Week's Differentiation Problem?

[Ackbach]

Here is this week's POTW:

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A student forgot the Product Rule for differentiation and made the mistake of thinking that $(fg)'=f'g'$. However, he was lucky and got the correct answer. The function $f$ that he used was $f(x)=e^{x^2}$ and the domain of his problem was the interval $\left(\tfrac12, \infty\right)$. What was the function $g$?

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[Ackbach]

Congratulations to kiwi and Aryth for their correct solutions! Kiwi's solution follows:

Spoiler

$$f'=2xf$$

$$(fg)'=g'f+gf'=g'f+2xfg=(2xg+g')f$$

$$f'g'=2xg'f$$

So to give the correct answer we require:

$$(fg)'=f'g'$$

$$\therefore (2xg+g')f=2xg'f$$

$$\therefore \frac {g'}g=\frac{2x}{2x-1}$$

$$\therefore \ln (g)=\int \frac{2x}{2x-1}dx$$

$$\therefore \ln (g)= \int 1 + \frac{1}{2x-1}dx=\frac 12(2x+\ln(2x-1))+C=x+\ln(\sqrt{2x-1}))+C$$

$$\therefore g= Ae^x \sqrt{2x-1}$$