Multi-variable differential question

[steejk]

Homework Statement

F = 1/2.a(T-Tc)M^2 + 1/4.bM^4

I need to find dF/dM

a,Tc,b are positive constants

2. The attempt at a solution

I assume this is to do with partial derivatives etc.

So I found:

∂F/∂T = 1/2.aM^2

∂F/∂M = TaM - TcaM + bM^3

And using a chain rule:

dF/dM = ∂F/∂T.dT/dM + ∂F/∂M.dM/dM

But not sure how to find dT/dM

 

[LCKurtz]

Homework Statement

F = 1/2.a(T-Tc)M^2 + 1/4.bM^4

I need to find dF/dM

Is that supposed to be$$
F=(\frac 1 2)a(T-T_c)M^2 +(\frac 1 4)bM^4$$

a,Tc,b are positive constants

2. The attempt at a solution

I assume this is to do with partial derivatives etc.

So I found:

∂F/∂T = 1/2.aM^2

∂F/∂M = TaM - TcaM + bM^3

And using a chain rule:

dF/dM = ∂F/∂T.dT/dM + ∂F/∂M.dM/dM

But not sure how to find dT/dM

The statement of the problem says find $\frac{dF}{dM}$. Isn't the expression just a polynomial in $M$? Hold everything else constant and differentiate it.

 

[steejk]

Is that supposed to be$$
F=(\frac 1 2)a(T-T_c)M^2 +(\frac 1 4)bM^4$$

Yes that's the correct equation.

Isn't T also a variable or have I just forgotten how to do basic differentiation?

 

[LCKurtz]

Homework Statement

F = 1/2.a(T-Tc)M^2 + 1/4.bM^4

I need to find dF/dM

a,Tc,b are positive constants

Yes that's the correct equation.

Isn't T also a variable or have I just forgotten how to do basic differentiation?

You said it was a constant above. And if it weren't, you would still calculate the partial derivative $\frac{\partial F}{\partial M}$ the same way.

 

[steejk]

You said it was a constant above. And if it weren't, you would still calculate the partial derivative $\frac{\partial F}{\partial M}$ the same way.

Sorry, Tc is a constant but not T.

I can see the partial derivative would hold everything else constant, but I need to find the total derivative dF/dM.

 

[LCKurtz]

I guess I misunderstood what you wanted. If T depends on M, then the chain rule in your original post would be correct. But without more information, I don't see you you could calculate $\frac{dT}{dM}$.