Question

Use the definition of the derivative to differentiate v=4/2 pie r^3

Answer 1

I suspect 4/2 should actually be 4/3, since 4/2 = 2, while 4/3 would make V the volume of a sphere with radius r. But I'll stick with what's given:

$\displaystyle \frac{dV}{dr} = \lim_{h\to0} \frac{2\pi(r+h)^3-2\pi r^3}{h}$

$\displaystyle \frac{dV}{dr} = 2\pi \lim_{h\to0} \frac{(r^3+3r^2h+3rh^2+h^3)- r^3}{h}$

$\displaystyle \frac{dV}{dr} = 2\pi \lim_{h\to0} \frac{3r^2h+3rh^2+h^3}{h}$

$\displaystyle \frac{dV}{dr} = 2\pi \lim_{h\to0} (3r^2+3rh+h^2)$

$\displaystyle \frac{dV}{dr} = 2\pi \cdot 3r^2 = \boxed{6\pi r^2}$

In Mathematica, you can check this result via

D[4/2*Pi*r^3, r]