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My math teacher recently asked my class to find if there are any functions that comply to the rule $$(f\cdot g)^\prime = f^\prime \cdot g^\prime$$ I have searched the web for an answer but I couldn't find it, could anyone help me with that? or at least point me to the right direction?

Batominovski
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Simon Kharmatsky
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  • Constant functions. – David Mitra Dec 15 '18 at 18:46
  • $f=0$ or $g=0$. – obscurans Dec 15 '18 at 18:54
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    You can expand the LHS to $f'g+fg'$. If this needs to be equal to the RHS then $f'g+fg'-f'g'=0$. If you group terms then $f'(g-g')+fg'=0$. Thus, if $f=0$ then $g$ can be anything, and if $g=0$ then $f$ can be anything. As already noted, if $f$ and $g$ are both constant of any value then our equation is still satisfied. Can there be any more solutions? If $g(x)=e^{x/2}$ and $f(x)=e^{-x}$ we still have a solution! I found all of these just by playing with the equation, so there could still be more. – RandomMathDude Dec 15 '18 at 18:56
  • @RandomMathDude You can do much more than thet. If $g$ is a given function, this is a linear first order differential equation in $f$, whose solution is known. – Jean-Claude Arbaut Dec 15 '18 at 19:02
  • What's * between functions? – Nosrati Dec 15 '18 at 19:06
  • @Jean-ClaudeArbaut Absolutely. I had assumed (perhaps erroneously) that this student is in a calculus class, and the exercise was intended to shed light on the Leibniz rule. – RandomMathDude Dec 15 '18 at 19:08
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    When the teacher said "find", did she/he really mean "find on the web"? – Blue Dec 15 '18 at 19:18

3 Answers3

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If $(fg)'=f'g'$, then $f'g+g'f=f'g'$, so $$(g-g')f'+g'f=0.$$ If $g=0$, then $f$ can be any differentiable function. If $g=g'$ and $g\neq 0$, then $g(x)=ae^x$ for some constant $a\neq 0$. Thus, $f=0$.

Suppose now that $g\neq g'$. Then, write $$f'+\frac{g'}{g-g'}f=0.$$ Therefore, $$(\mu f)'=0,$$ where $$\mu(x)=\exp\left(\int\frac{g'(x)}{g(x)-g'(x)}dx\right).$$ Thus, for some constant $b$, $$f(x)=\frac{b}{\mu(x)}=b\exp\left(-\int\frac{g'(x)}{g(x)-g'(x)}dx\right).$$

For example, if $g(x)=x$, then we can take $\mu(x)=x-1$. That is, $f(x)=\frac{b}{x-1}$. For another example, if $g(x)=\sin x$ for $x\in(-\pi/4,\pi/4)$, then we can take $\mu(x)=e^{-x/2}\sqrt{\cos x-\sin x}$, so that $f(x)=\frac{e^{x/2}}{\sqrt{\cos x-\sin x}}$. For one last example, if $g(x)=\sqrt{x^2+1}$, then we can take $\mu(x)=\sqrt{x^2-x+1}\exp\left(\frac{\arctan\left(\frac{2x-1}{\sqrt3}\right)}{\sqrt{3}}\right)$ and so $f(x)=\frac{b}{\sqrt{x^2-x+1}}\exp\left(-\frac{\arctan\left(\frac{2x-1}{\sqrt3}\right)}{\sqrt{3}}\right)$.

  • I had the intention, but not the patience. Well done sir, a complete answer. – Lucas Henrique Dec 15 '18 at 21:02
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    There is another infinite family of solutions apart from hamam_Abdallah's family. It contains your example with $g(x)=x$. This family consists of $(f,g)$ such that there exists $\alpha\in\mathbb{R}$ for which $$g(x)=x^\alpha\text{ and }f(x)=\left(x-\alpha\right)^{-\alpha}$$ with the domain $\mathbb{R}\setminus{0}$ if $\alpha<0$, $\mathbb{R}$ if $\alpha=0$, and $\mathbb{R}\setminus{\alpha}$ if $\alpha>0$. – Batominovski Dec 16 '18 at 12:35
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if $f(x)=e^{ax}$ and $g(x)=e^{bx}$ then

$a+b=ab$. so we can take

$$f(x)=e^{ax}$$ and $$ g(x)=e^{\frac{ax}{a-1}}$$

hamam_Abdallah
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$(0\cdot 0)' = \\ 0' = \\ 0'\cdot0'$

Lucas Henrique
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