When do the Freshman's dream product and quotient rules for differentiation hold? This is motivated by looking at the calculus exams of some of my undergraduate students. A recurring mistake is assuming that the derivative of the product of functions is a product of derivatives and the derivative of the quotient of two functions is the quotient of their derivatives. 
This might be an ill formed question, but do their exist a pair of functions such that these rules hold. Explicitly, do there exist $f(x),g(x)$ satisfying all of:
\begin{align*}
f&\neq 0;\\
\frac{d(fg)}{dx}&=\frac{df}{dx}\cdot \frac{dg}{dx}\\
\text{and }\frac{d(f/g)}{dx}&=\frac{df}{dx}/ \frac{dg}{dx}\text{ ?}
\end{align*}
EDIT: Just to clarify my question, looking at the responses so far. I am looking for functions satisfying all three conditions above simultaneously.
 A: The product rule equality gives $f'g+g'f = f'g'$, or $-fg' = f'g-f'g'$. From this, using the quotient rule equality we get
$$
\frac{f'}{g'} = \frac{gf'-fg'}{g^2} = \frac{gf' +f'g - f'g'}{g^2}
= f'\left(\frac{2g-g'}{g^2}\right).
$$
If $f$ is not constant, then $f'\neq 0$. So we get $\frac{1}{g'} = \frac{2g-g'}{g^2}$, or $g^2 - 2gg'+(g')^2 = 0$, or $(g-g')^2 = 0$; hence $g=g'$.
Therefore, from the product rule again we have $f'g'=f'g+fg'$, or $f'g = f'g+fg$, so $fg=0$. This requires $g=0$, which makes the quotient rule impossible.
If $f$ is constant, then the product rule equality requires $g$ constant, and then the quotient rule equality cannot hold. 
A: Another answer to your question from a different angle is this: They hold in a different kind of calculus.
If you swap subtraction for division and division for taking roots in the usual definition of the derivative, you get 
$$f^*(x) = \lim_{\Delta x \to 0} \left(\frac{f(x+\Delta x)}{f(x)}\right)^{\frac{1}{\Delta x}}.$$
It's not too hard to prove that $(fg)^* = f^* g^*$ and $(f/g)^* = f^*/g^*$ under this definition.
The calculus that results from this definition of the derivative (and the corresponding definition of the integral) goes by various names, such as "multiplicative calculus," "non-Newtonian calculus," and "product calculus."  Many of the standard results in the usual calculus (e.g., Fundamental Theorem, Mean Value Theorem, l'Hopital-type rules, Taylor's Theorem) can be redone for this multiplicative calculus.  The idea goes back at least to Vito Volterra in the late 1800s.
According to the Wikipedia page, "Opinions differ as to the usefulness of the multiplicative calculi," but there are some applications.  For instance, the product derivative measures the multiplicative rate of change of some function and so is useful when you're interested in looking at, say, growth rates of stock prices.  The product integral has some corresponding applications; my favorite is that it can be used to calculate geometric means (in the same way that the usual integral can find arithmetic means).
Multiplicative calculus also has a close relationship to the usual calculus.  For instance, the derivatives are related via 
$$f^*(x) = \exp\left(\frac{d}{dx}\ln |f(x)|\right) = e^{f'(x)/f(x)}.$$
The product integral and the usual integral have a similar relationship.  In fact, I think part of the reason some folks don't find this multiplicative calculus all that interesting is that product derivatives and integrals can so easily be expressed in terms of the usual ones.  The question, I suppose, is whether there is really anything new and different going on here.  (Note also that this relationship I just gave means that the product derivative is just $\exp$ of the logarithmic derivative of $f$.)
Anyway, a few years ago I wrote a survey paper and had a student do a summer research project on this before I realized how much was already out there and before the Wikipedia page appeared.  There are lots of other references and information on the Wikipedia page, so if you find this interesting you should check it out.  The page on the product integral is also a good source.
A: I don't think there is a solution.
$f'g'=f'g+g'f$ implies $\frac{f'}{f}=\frac{g'}{g'-g}$ where the denominators are nonzero.
$\frac{f'}{g'}=\frac{f'g-g'f}{g^2}$ implies $\frac{f'}{f}=\frac{(g')^2}{g(g'-g)}$ where the denominators are nonzero.  
If $f$ and $g$ are nonzero and $g'-g$ is nonzero, this implies that $\frac{g'}{g'-g}=\frac{(g')^2}{g(g'-g)}$, which implies $g'=g$.  Thus $g=g'$ if $f$ and $g$ are nonzero, so $g(x)=ce^x$.  But now from $f'g'=f'g+g'f$ we have $ce^xf(x)=0$, which implies $f=0$.

For solutions to just the product rule, the equations above suggest taking $g$ such that $\frac{g'}{g'-g}$ is integrable, and 
$$f(x)=\exp\left(\int_a^x\frac{g'(t)}{g'(t)-g(t)}dt\right),$$ which in particular works for $f(x)=e^{bx}$ and $g(x)=e^{cx}$ with $bc=b+c$ (as seen in a now deleted answer).  Similarly, for just the quotient rule, one could try finding $g$ such that $\frac{(g')^2}{g(g'-g)}$ is integrable, and then take 
$$f(x)=\exp\left(\int_a^x\frac{g'(t)^2}{g(t)(g'(t)-g(t))}dt\right).$$  
For example, $f(x)=e^{bx}$ and $g(x)=e^{cx}$ with $bc=b+c^2$.
