A very basic question about the derivative of a quotient I have a quotient function of following type $$h(x)=\frac{f(x)}{g(x)}$$ I know that $f(x)$ increases with $x$ and $g(x)$ decreases with $x$ so can we conclude that the derivative of $h(x)$ is positive?
Thanks in advance
 A: The quotient rule states:
$$h'(x) = {g(x)f'(x) - f(x)g'(x) \over g(x)^2}$$
Following your comments, we are given that $f'(x) > 0$ and $g'(x) < 0$, additionally that $f(x) < 0$ and $g(x) > 0$, you can see that:
$$g(x)f'(x) > 0$$
And:
$$f(x)g'(x) > 0$$
Because a negative multiplied by a negative results in a positive. Since we're subtracting a positive from a positive, we still are not able to make a definite conclusion about $h'(x)$.
A: If 
$f(x) = x^3, \tag 1$
$g(x) = - x^3, \tag 2$
then $f(x)$ is strictly increasing, $g(x)$ is strictly decreasing, but on any interval where $x \ne 0$
$h(x) = \dfrac{f(x)}{g(x)} = \dfrac{x^3}{-x^3} = -1, \tag 3$
which is a constant, so 
$h'(x) = 0.  \tag 4$
Thus we have a counter-example.
I think it's worth pointing out that if both 
$f(x), g(x) > 0 \tag 5$
then
$h'(x) = \left ( \dfrac{f(x)}{g(x)} \right )' = \dfrac{f'(x) g(x) - f(x)g'(x)}{(g(x))^2} > 0. \tag 6$
For example, if instead of (2) we take
$g(x) = x^{-3}, \tag 6$
then
$h(x) = x^6, \tag 7$
$h'(x) = 6x^5 > 0, \; x > 0. \tag 8$
Furthermore, if
$f(x), g(x) < 0, \tag 9$
then 
$h'(x) < 0 \tag{10}$
as well.
