A very basic question about the derivative of a quotient

Question

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

Answer 1

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 $h'(x)$.

Answer 2

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.