use the definition of the derivative to evaluate v=4/2pir^3
f′(x)=limh→0f(x+h)−f(x)hf'(x)=\displaystyle{\lim_{h\to 0}}\frac{f(x+h)-f(x)}{h}f′(x)=h→0limhf(x+h)−f(x)
dvdr=4π2limh→0v(r+h)−f(r)h=2πlimh→0(r+h)3−r3h=\frac{dv}{dr}=\frac{4\pi}{2}\displaystyle{\lim_{h\to 0}}\frac{v(r+h)-f(r)}{h}=2\pi \displaystyle{\lim_{h\to 0}}\frac{(r+h)^3-r^3}{h}=drdv=24πh→0limhv(r+h)−f(r)=2πh→0limh(r+h)3−r3=
=2πlimh→0r3+3r2h+3rh2+h3−r3h=2πlimh→0(3r2+3rh+h2)==2\pi \displaystyle{\lim_{h\to 0}}\frac{r^3+3r^2h+3rh^2+h^3-r^3}{h}=2\pi \displaystyle{\lim_{h\to 0}}(3r^2+3rh+h^2)==2πh→0limhr3+3r2h+3rh2+h3−r3=2πh→0lim(3r2+3rh+h2)=
=2π⋅3r2=6πr2=2\pi\cdot 3r^2=6\pi r^2=2π⋅3r2=6πr2
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