A way to think about differentiation by thinking of functions as constants We are all familiar with the product rule for differentiating: $$[f(x)g(x)]'=f(x)'g(x)+g(x)'f(x)$$ This result can be achieved if we think of $g(x)$ as a constant and differentiate $f(x)$ and then think of $f(x)$ as a constant and differentiate $g(x)$ and then add the two (if you see what I mean?). The same thing happens with the quotient rule: $$\left[\frac{f(x)}{g(x)}\right]'=\frac{f'(x)}{g(x)}-\frac{f(x)g'(x)}{(g(x))^2}=\frac{f'(x)g(x)-f(x)g'(x)}{(g(x))^2}$$. We can also use this way of thinking to differentiate $f(x)^{g(x)}$:
$$\left[f(x)^{g(x)}\right]=g(x)f(x)^{g(x)-1}f'(x)+f(x)^{g(x)}\ln(f(x))g'(x)$$
Can anyone explain why this works (the idea of treating one function as a constant and differentiating the other and then adding it to the derivative of the other function with the first function as a constant)? It reminds me of vector addition because we are somehow splitting the differential but I do not really understand it at all. 
 A: For the case of $f(x)g(x)$, what you are doing is defining a new function $F : R^2 \to R$ with $F(x,y) = f(x)g(y)$. The derivative of this can be computed by taking partials with respect to $x$ and $y$, so it is $(f'(x)g(y), f(x)g'(y))$. Next, you set $x = y$, which means that you are considering the value of this derivative in the direction $(1,1)$, so that you have $(f'(x)g(y), f(x)g'(y)) (1,1) = f'(x)g(y) + f(x)g'(y)$. Note that the function $F$ is $f(x)g(x)$ along the line $x = y$ in $R^2$, and that its directional derivative along this line is the same as the derivative of $f(x)g(x)$ with respect to $x$. (Since moving 1 unit along the $x$ direction would require moving 1 unit along the $y$ direction.)
The other cases follow the same train of thought - you are thinking of a product or a quotient as a multivariable function and finding the directional derivative.
For the case $f(x)^{g(y)}$ you have to be careful since this function may not always be well defined.
A: What is at stake here is linearity of differentiation, the chain rule, and maybe other subtle stuff. Expressions of the form
$$x\mapsto f(x)\cdot g(x), \quad x\mapsto{f(x)\over g(x)},\quad x\mapsto f(x)^{g(x)}\tag{1}$$
can be considered as compositions of the maps
${\bf h}:\ x\mapsto\bigl(f(x),g(x)\bigr)$ and $p: (u,v)\mapsto u\cdot v\ $:
$$f(x)\cdot g(x)=p\circ{\bf h}(x)\ ,$$
and similarly for the other two in $(1)$. One has $${\bf h}'(x)=\bigl(f'(x),g'(x)\bigr), \qquad \nabla p(u,v)=(v,u),$$
and therefore by the chain rule
$${d\over dx}\bigl(f(x)\cdot g(x)\bigr)={d\over dx}(p\circ{\bf h})(x)=\nabla p\bigl({\bf h}(x)\bigr)\cdot{\bf h}'(x)=g(x)f'(x)+f(x)g'(x)\ .$$
Arguing similarly for $f/g$ and $f^g$ one arrives at the corresponding correct formulas. (When multivariable calculus is not available these formulas are of course proven by ad hoc arguments.)
A: For a good answer, there was an article posted in Vol. 35, No. 5 of "The college mathematics journal" published by the MAA written by Noah Samuel Brannen and Ben Ford on this very topic. In fact, here is a link:
http://www.maa.org/sites/default/files/brannen11200427315.pdf
