Analytically solving for T in $\dfrac{d}{dz} (\dfrac{dT}{dz})=0$?

Question

So as the title says, I'm trying to solve $\dfrac{d}{dz} (\dfrac{dT}{dz})=0$?

I've been given the result $\dfrac{dT}{dz}=c$ where c is just a constant and the only way I can see of obtaining this result is this way (NB I've heard mathematicians hate it when people treat dy/dx etc. like fractions..)

$\dfrac{dT}{dz}=0 \times \int dz = \int 0 \ dz=c$

in the same way that: $\int 8+0 \ dx = \int 8dx+\int0 \ dx$ is $8x+c$?

What I'm confused about is why in this case, it seems that $0 * \int dz\neq0$.

Can anyone shed some light on this?

Answer 1

This follows from the following fact: if $f' = 0$ then $f$ is a constant function.