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 $(a,b)$.
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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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