Derivatives with multiple variable, help

[jonhendrix86]

Homework Statement

This is an Optimization Problem, find the maximum value.

P(R)=(E^2*R)/(R+r)^2

Homework Equations

P'(R)=?

The Attempt at a Solution

I have the solution to this problem, and I can solve it, I just don't understand some parts. I tend to think that the derivative of E^2*R = 2ER, like the power rule similar to if I solved x^2*y I would get 2xy. But the derivative of E^2*R is just E^2 and I cannot figure it out. Can someone please explain to me with mathematical proof why the derivative of E^2*R=E^2? Thanks a bunch.

 

[Dick]

Homework Statement

This is an Optimization Problem, find the maximum value.

P(R)=(E^2*R)/(R+r)^2

Homework Equations

P'(R)=?

The Attempt at a Solution

I have the solution to this problem, and I can solve it, I just don't understand some parts. I tend to think that the derivative of E^2*R = 2ER, like the power rule similar to if I solved x^2*y I would get 2xy. But the derivative of E^2*R is just E^2 and I cannot figure it out. Can someone please explain to me with mathematical proof why the derivative of E^2*R=E^2? Thanks a bunch.

It's because E and r are constants. R is the variable. The derivative of a constant is zero. If c is a constant then d/dR(cR)=c.

 

[haruspex]

P(R)=(E^2*R)/(R+r)^2
I tend to think that the derivative of E^2*R = 2ER, like the power rule similar to if I solved x^2*y I would get 2xy. But the derivative of E^2*R is just E^2 and I cannot figure it out. Can someone please explain to me with mathematical proof why the derivative of E^2*R=E^2? Thanks a bunch.

E is a constant here, yes? So E^2 is as well. You can just take the constant outside the derivative:
d/dx(c f(x)) = c d/dx(f(x)).
Note that this is not some special treatment of constants. You can get the same result using the product rule:
d/dx(c f(x)) = c d/dx(f(x)) + f(x) dc/dx, but because c is a constant dc/dx = 0.

 

[jonhendrix86]

Thank you! I see it now.

Concerning E^2*R,

if y'=uv'+vu', then y'=(E^2)(1)+(R)(0)=E^2.