# 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 $$(a,b)$$.

• Looks correct to me. Commented Nov 13, 2020 at 1:19
• 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 $C'$? Commented Nov 13, 2020 at 1:19
• Maybe I'll use lowercase c and uppercase C. Commented Nov 13, 2020 at 1:20
• Best to keep your constants uppercase. Perhaps $C$ and $D$ or $C_1$ and $C_2$. Commented Nov 13, 2020 at 1:55

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.