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 $(\frac{1}{2} , \infty)$What was the function $g$?

For the product rule

$(fg)'=f'g + fg'=e^{x^2}\cdot 2x\cdot g+e^{x^2}\cdot g'$

What the student did

$(fg)'=f'g'=e^{x^2}\cdot 2x \cdot g'$

Now we assume both of these to be equivalent to other

$$(fg)' =(fg)'$$

$$ e^{x^2}\cdot 2x\cdot g+e^{x^2}\cdot g'=e^{x^2}\cdot 2x \cdot g'$$

$$ e^{x^2}\cdot 2x\cdot g = g' (e^{x^2} \cdot 2x - e^{x^2} )$$

$$ \frac{e^{x^2}\cdot 2x}{e^{x^2} \cdot 2x - e^{x^2}} = \frac{g'}{g} $$

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

This is the part I am unsure about since I'm not sure if I can cancel the $e^{x^2} out $

After dominiks comment I continue

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

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

$$ \int 1 + \frac{1}{2x-1} dx = \int \frac{g'}{g} dx$$

$$ x+\frac{\ln|2x-1|}{2}+c =\ln|g|$$

$$ g=e^{x+\frac{\ln|2x-1|}{2}+c}$$

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

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    $\begingroup$ Your calculations are correct. Now you just need to solve the differential equation. $\endgroup$ – Dominik Dec 16 '16 at 23:16
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    $\begingroup$ Of course you can cancel $e^{x^2}$ out; you're dividing the top and bottom by a non-zero function. $\endgroup$ – Omnomnomnom Dec 16 '16 at 23:18
  • $\begingroup$ @Dominik Do you mind checking my work now? $\endgroup$ – bigfocalchord Dec 16 '16 at 23:21
  • $\begingroup$ @Omnomnomnom Thank you for the reassurance! $\endgroup$ – bigfocalchord Dec 16 '16 at 23:21
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    $\begingroup$ You start off correct, but have a small mistake on the last line. The logarithm shouldn't be there. Also, note that you don't need absolute values around $2x -1$, as $x > \frac{1}{2}$. $\endgroup$ – Dominik Dec 16 '16 at 23:25

As stated in the comments, your cancellation of $e^{x^2}$ was fine. However, you've made a mistake finding $g$. You should find $$ g=e^{x+\frac{\ln|2x-1|}{2}+c} = e^ce^x\left[e^{\ln|2x - 1|} \right]^{1/2} = Ae^{x} \sqrt{|2x-1|} $$ and, since $x > 1/2$, the absolute value is redundant.


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