Are There Functions Where $(f\cdot g)^\prime$ is equal to $f^\prime \cdot g^\prime$?

Question

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?

Answer 1

if $f(x)=e^{ax}$ and $g(x)=e^{bx}$ then

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

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

Answer 2

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

Answer 3

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 $f(x)=\frac{b}{\sqrt{x^2-x+1}}\exp\left(-\frac{\arctan\left(\frac{2x-1}{\sqrt3}\right)}{\sqrt{3}}\right)$.