How do you get m=0? from taking the second derivative? does c=1?

Can you check my work on how I get the derivative of R=M^2(c/2-m/3) NEXT R=1/2(CM^2)-1/3(M^3) NEXT dR/dM=CM-M^2. I checked in the back, and that was the answer for the first part of the equation. Isn't the derivative of constant c, zero? I am confused.

You have it correct. yes, the derivative of a constant is zero, but you didn't take that derivative.

if y= ax
dy/dx= a

If you took the long way, the uv rule..

y=ax
dy/dx= a dx/dx + x da/dx
but da/dx is zero (derivative of a constant is zero), os
dy/dx= a

To find the second derivative, you need to take the derivative of the first derivative. But before diving into the specific question, let's clarify some concepts.

When you have a constant value, such as "c" in your equation, its derivative with respect to any variable is always zero. Therefore, if you have R = M^2(c/2 - m/3), the derivative of c/2 with respect to M is zero, and the derivative of -m/3 with respect to M is also zero. So, in this case, there is no term involving c or m in the derivative with respect to M.

Now let's check your work on finding the derivative of R = M^2(c/2 - m/3):

Step 1: R = M^2(c/2 - m/3)
Step 2: Differentiate each term with respect to M:
dR/dM = d/dM [M^2(c/2 - m/3)]
Step 3: Apply the power rule:
dR/dM = 2M^(2-1)(c/2 - m/3)
dR/dM = 2M(c/2 - m/3)
dR/dM = CM - 2Mm/3

So, your work is correct up to this point. The derivative of R with respect to M is indeed CM - 2Mm/3.

To find the second derivative, we need to take the derivative of the first derivative. Taking the derivative of CM - 2Mm/3 with respect to M:

Step 1: Differentiate each term with respect to M:
d^2R/dM^2 = d/dM [CM - 2Mm/3]
Step 2: Apply the power rule:
d^2R/dM^2 = 0 - 2m/3
d^2R/dM^2 = -2m/3

So, the second derivative of R with respect to M is -2m/3.

Regarding your question about m = 0, we haven't specifically assumed or proven that m equals zero in the given equation. If you have additional information or conditions that suggest m is zero, please include that information for a more accurate response.