It is known that the volume of a sphere with radius R is v = 4 / 3 π R ^ 3. By using the definition of derivative, it is proved that the surface area of the sphere is s = 4 π R ^ 2

It is known that the volume of a sphere with radius R is v = 4 / 3 π R ^ 3. By using the definition of derivative, it is proved that the surface area of the sphere is s = 4 π R ^ 2

ΔV=f(r+Δr)-f(r)=4/3π(r+Δr^)-3,4/3πr^3=4/3π[(r+Δr)^3-r^3]=4/3π(3r^2*Δr+3r*Δr^2+Δr^3)
s=limΔV/Δr=4/3π(3r^2+3r*Δr+Δr^2)=4πr^2
Δr->0
It's over

Why is the surface area of the sphere (4 π R ^ 2) exactly the derivative of the sphere volume (4 / 3 π R ^ 3)?
Including the circumference of the circle (2 π R) is exactly the derivative of the area of the circle (π R ^ 2)
By chance, I found this rule in it,

Proof: let's first prove that the circumference of a circle (2 π R) is exactly the area of a circle (π R ^ 2) with respect to the derivative of R. let's suppose that the radius of one circle is r, and the radius of the other circle concentric with it is R + △ R. let's first look at the circle band composed of two concentric circles, and its area is π (R + △ R) ^ 2 - π R ^ 2

Given the function v (R) = 4 / 3 π R ^ 3 of the volume of a sphere with respect to its radius, its derivative V '(R) = 4 π R ^ 2 is just the surface area of the sphere
The formula is

The area of the circle s (R) = π R ^ 2, the circumference of the circle s' (R) = 2 π R is just right. I can't help but think about it and come up with this