Why is $\pi$ not 0 when differentiated? I'm currently a student looking through the differentiation topic and came across this problem as follows...
$$V=125\pi r-\pi r^3$$
Where   $ r$ is a variable and $r^3$ represents r cubed.
The first derivative $\frac {dV}{dr} = 125\pi-3r^2 \pi$
The second derivative $\frac {d^2V}{dr^2} = -6r\pi$
The formula and derivatives are both correct, however, my original attempt at them was the exact same apart from the exclusion of $\pi$.
If $y=57$ were to be differentiated the result would be $0$ according to Wallis' law. But as$\pi$ is simply a constant real number, why does it not simply disappear? (For lack of a better word.)

Let us check it out by looking at what the definition says. There is nothing mysterious.
By differentiating the function $V = 125\pi r - \pi r^{3}$ (where $\pi$ is a constant like $125$ so the value of $V$ is dependent on $r$ alone) at a point $r$ we mean taking the limit
$$
\lim_{h \to 0} \frac{[125 \pi (r+h) - \pi (r+h)^{3}] - [125 \pi r - \pi r^{3}]}{h}.
$$
Now you see that $\pi$, like $125$, does not contribute anything to the differentiation of this function. 
