Why is the derivative of a monotonic transformation of a utility function assumed to always be greater than 0?

Let's look at the use of monotonic transformations of utility functions (which I guess is the most frequent occurrence of this concept in econ).

Let $u: mathbb{R}^n_+ to mathbb{R}$ be a utility function. We say that $g: mathbb{R}^n_+ to mathbb{R}$ is a monotonic transformation of $u$ if for all $x, y in mathbb{R}^n_+$: $$ u(x) ge u(y) iff g(x) ge g(y). $$

Let $D subseteq mathbb{R}$. Then $h: D to mathbb{R}$ is called strictly increasing if for all $x, y in D$: $$ x ge y iff h(x) ge h(y). $$

There's the following result:

Th Let $D$ be the range of $u$, i.e. $D = u(mathbb{R}^n_+)$. Then $g$ is a monotonic transformation of $u$ iff there exists a strictly increasing function $h: D to mathbb{R}$ such that $g(x) = h(u(x))$.

proof: ($leftarrow$) If $g(x) = h(u(x))$ and $u(x) ge (>) u(y)$ then $h(u(x)) ge (>) h(u(y))$ so $g$ is indeed a monotone transformation of $u$.

($rightarrow$) For the reverse, define $$ h(z) = g(x) text{ whenever } u(x) = z. $$ First we need to check that $h$ is indeed a function. As such, let $x, y$ be such that $u(x) = u(y) = z$. We need to show that $g(x) = g(y)$. Indeed as $g$ is a monotone transformation, we have that $g(x) ge g(y)$ and $g(y) ge g(x)$ so $g(x) = g(y)$. To see that $h$ is strictly increasing, let $z ge (>) w$ and let $x$ and $y$ be such that $u(x) = z ge (>) w = u(y)$. Then as $g$ is a monotone transformation $h(z) = g(x) ge (>) g(y) = h(w)$. $blacksquare$

To see the connection with the derivatives we have the following:

Th if $g: D to mathbb{R}$ is differentiable, $D$ is convex and if $g'(x) > 0$ for all $x$ in $D$, then $g$ is strictly monotone.

proof: Let $x ge (>) y$ then by the mean value theorem there is a $c in [x,y]$ such that: $$ g(y) - g(x) = g'(c)(y - x) ge (>) 0. $$

Remark: as @Michael Greinecker said, the reverse is not true, there are strictly increasing functions whose derivative is zero at some points, like $g(x) = x^3$.

But how is that true for a negative monotonic transformation such as g(x)=−x where g′(x)=−1?

In principle we could define the reverse notions. Say that $tilde g$ is a "negative" monotonic transformation of $u$ if for all $x, y in mathbb{R}^n_+$: $$ u(x) ge u(y) iff tilde g(x) le tilde g(y). $$

Notice that $tilde g$ reverses the ranking given by $u$.

Next, we can look at functions $tilde h: D to mathbb{R}$ that are strictly decreasing: $$ x ge y iff tilde h(x) le tilde h(y). $$

We have the following:

Th The function $tilde g$ is a negative monotone transformation of $u$ iff there exists a strictly decreasing function $tilde h: D to mathbb{R}$ such that $tilde g(x) = tilde h(u(x))$.

The proof is similar to the proof for the monotone/strictly increasing case.

Remark: In economics, "negative" monotonic transformations are not really used. We would like the functions $u$ and $g$ to give the same ranking over all bundles $x in mathbb{R}^n_+$ then they should be monotone transformations of each other. If you use a negative monotonic transformation, you are reversing the order.