What is the derivative of $\frac{1}{f(x)g(x)}$? I feel like this is a very basic question, yet I struggle with it immensely. I know that $(f(x)*g(x)) = f(x)'g(x)+f(x)g(x)'$, but how to use that in order to figure out $\frac{1}{f(x)g(x)}$ ?
 A: Use the chain rule. Since
$$\frac{d}{dx}\frac{1}{x}=-\frac{1}{x^2}$$
and
$$\frac{d}{dx}f(x)g(x)=f'(x)g(x)+f(x)g'(x)$$
then
$$\frac{d}{dx}\frac{1}{f(x)g(x)}=-\frac{1}{f(x)^2g(x)^2}*(f'(x)g(x)+f(x)g'(x))=-\frac{f'(x)g(x)+f(x)g'(x)}{f(x)^2g(x)^2}$$
A: There are a couple of ways to think about this; but, perhaps the easiest to start with would be to realize that
$$
\frac{1}{f(x)g(x)}=\frac{1}{f(x)}\cdot\frac{1}{g(x)}.
$$
So, by the product rule,
$$
\frac{d}{dx}\left[\frac{1}{f(x)}\cdot\frac{1}{g(x)}\right]=\frac{1}{f(x)}\cdot\frac{d}{dx}\left[\frac{1}{g(x)}\right]+\frac{1}{g(x)}\cdot\frac{d}{dx}\left[\frac{1}{f(x)}\right].
$$
From here, you can use the quotient rule (or the chain rule) to evaluate these two derivatives.
A: rewriting as $$(f(x)g(x))^{-1}$$ we get by the power and chain rule
$$(-1)(f(x)g(x))^{-2}(f'(x)g(x)+f(x)g'(x))$$
if $$f(x)g(x)\ne 0$$
A: take logarithms , now you have
$$ lny= -lnf(x) -lng(x)$$ so the derivative is
$$ \frac{y'(x)}{y(x)}= \frac{-f'(x)}{f(x)}-\frac{g'(x)}{g(x)} $$
now multiply both sides by $ y(x) $
A: Think of $\frac{1}{f(x)g(x)}$ as $(f(x)g(x))^{-1}$.  Then you can use the chain rule to differentiate:
\begin{split}[(f(x)g(x))^{-1}]' &= -1(f(x)g(x))^{-1 - 1} \cdot [f(x)g(x)]' \\ &= -1(f(x)g(x))^{-2}\cdot[f'(x)g(x) + f(x)g'(x)] \\ &=- \frac{f'(x)g(x) + f(x)g'(x)}{(f(x)g(x))^{2}}\end{split}
