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Question: A not uncommon calculus mistake is to believe that the product rule for derivatives says that $(fg)'=f'g'$. If $f(x)=e^{x^2}$, determine, with proof, whether there exists an open interval $(a,b)$ and a nonzero differentiable function $g$ defined on $(a,b)$ such that this wrong product rule is true for $x$ in $(a,b)$.

Is my solution correct? Is there anything I need to improve?

My solution:

We're trying to find a function $g$ such that $(fg)'=f'g'.$ Using the product rule, we get $$f'g + fg' = f'g'.$$ Plugging in both $f(x)=e^{x^2}$ and $f'(x)=2xe^{x^2},$ we get $$2xe^{x^2}g + e^{x^2}g' = 2xe^{x^2}g'.$$ Simplifying and canceling $e^{x^2},$ we get $$\frac{g'}{g} = \frac{2x-1}{2x} = 1 + \frac{1}{2x-1}.$$ Taking the integral of both sides, $$\int \frac{dg}g = \int 1+ \frac{1}{2x-1} \, dx.$$ $$\log|g| = \frac{\log|2x-1|}{2} + x + C.$$ $$|g| = e^{\frac{\log|2x-1|}{2}}e^xe^C.$$ Letting $e^C=K,$ we get $$\boxed{g = Ke^x\sqrt{|2x-1|}},$$ Where K is a constant greater than 0. Therefore, on any interval $(a,b)$ that does not contain the value of $\frac{1}{2},$ there exists nonzero differentiable function $g$ defined on $(a,b)$ such that $(fg)'=f'g'$ is true for $x$ in $(a,b)$.

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    $\begingroup$ Looks correct to me. $\endgroup$ Commented Nov 13, 2020 at 1:19
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    $\begingroup$ Looks pretty good. The one thing I think I would improve is you've bound $C$ twice: first as the arbitrary constant of indefinite integration, and second as the result of $e^C$, since $C$, being an arbitrary constant, means $e^C$ is also an arbitrary constant. You might want to call $e^C$ as something other than $C$ again, maybe $C'$? $\endgroup$
    – Andrew L
    Commented Nov 13, 2020 at 1:19
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    $\begingroup$ Maybe I'll use lowercase c and uppercase C. $\endgroup$
    – CSS Jowoo
    Commented Nov 13, 2020 at 1:20
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    $\begingroup$ Best to keep your constants uppercase. Perhaps $C$ and $D$ or $C_1$ and $C_2$. $\endgroup$ Commented Nov 13, 2020 at 1:55

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I think you have gone far beyond the requirements of the question, which asked merely to prove that there exists an open interval and a non-zero differentiable function with the stated properties.

To prove that such an interval and function exist, you can simply choose an open interval -- for example, $(a,b) = (1,2)$ -- and one function on that interval, for example $g(x) = e^x \sqrt{2x - 1}.$ Then show that $f(x)$ and $g(x)$ (and therefore also $(fg)(x)$) are defined on your chosen interval and calculate $f'$, $g'$, and $(fg)'$:

\begin{align} f'(x) &= \frac{\mathrm d}{\mathrm dx} e^{x^2} = 2x e^{x^2}, \\ g'(x) &= \frac{\mathrm d}{\mathrm dx}\left(e^x \sqrt{2x - 1}\right) = \frac{2x e^x}{\sqrt{2x - 1}}, \\ (fg)'(x) &= \frac{\mathrm d}{\mathrm dx} \left(e^{x^2}\cdot e^x \sqrt{2x - 1}\right) \\ &= \frac{4 x^2 e^{x^2 + x}}{\sqrt{2x - 1}} \\ &= 2x e^{x^2} \cdot \frac{2x e^x}{\sqrt{2x - 1}} \\ &= f'(x)\cdot g'(x). \end{align}

Note that it was not necessary to take the absolute value of $2x-1$ in any of these formulas, since $2x - 1 > 0$ everywhere in the domain of $f$ and $g$ as defined in this proof.

And that is essentially the proof of existence of an interval and a function with the desired properties. You could choose the interval $\left(\frac12,\infty\right)$ rather than $(1,2)$ for the existence proof (as I would) since it is one of the two maximal intervals on which such a function $g$ is defined, and perhaps you might like to show off your clever method of identifying the entire family of functions that might be the desired function $g$ (as I likely would), but all of that is bonus material beyond the plain meaning of the question.

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