Is my Calculus Proof Correct? 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 $(a,b)$.
 A: 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.
