Question #265864

use the definition of the derivative to evaluate v=4/2pir^3


1
Expert's answer
2021-11-15T16:32:47-0500

f(x)=limh0f(x+h)f(x)hf'(x)=\displaystyle{\lim_{h\to 0}}\frac{f(x+h)-f(x)}{h}


dvdr=4π2limh0v(r+h)f(r)h=2πlimh0(r+h)3r3h=\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}=


=2πlimh0r3+3r2h+3rh2+h3r3h=2πlimh0(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)=


=2π3r2=6πr2=2\pi\cdot 3r^2=6\pi r^2


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