# Converting partial derivative w.r.t. T to partial derivative w.r.t. 1/T

1. Apr 1, 2008

### knulp

Hi, I have a question about a certain step in the following problem/derivation, which you'll see in square brackets:

Show that T * ($$\partial$$/ $$\partial$$T) = (-1/T) * ($$\partial$$/ $$\partial$$(1/T))
["$$\partial$$/$$\partial$$T" is the operator that takes the partial derivative of something with respect to T]

Showing that this is true is a little tricky. For example, we can define F = 1/T. Then ($$\partial$$F/ $$\partial$$T) = -1/T^2 and ($$\partial$$F/ $$\partial$$F) = 1. So we can write

($$\partial$$F/ $$\partial$$T) = (-1/T^2) ($$\partial$$F/ $$\partial$$F).

[In the next step he drops the F, so it's now an operator for an arbitrary function, but still with respect to F… Is this really okay?]

($$\partial$$/ $$\partial$$T) = (-1/T^2)($$\partial$$/ $$\partial$$F)

= (-1/T^2) ($$\partial$$/ $$\partial$$(1/T)).

Multiplying by T, T($$\partial$$/ $$\partial$$T) = (-1/T)($$\partial$$/ $$\partial$$(1/T)) and we're done.

2. Apr 1, 2008

### Pere Callahan

The operator identity holds by definition if, when acting on suitable functions, both sides of the identity give the same result. So in my understanding it is not quite right to conclude the operator identiy from only having shown that when acting on F they give the same result. However you can easily cure this. It is just the chain rule. Let g(T) be some function, F(T) = 1/T,. then

$$\frac{\partial g}{\partial T} = \frac{\partial F}{\partial T}\frac{\partial g}{\partial F} = -\frac{1}{T^2}\frac{\partial g}{\partial F}$$

This is true for any suitable function g so

$$\frac{\partial }{\partial T} = -\frac{1}{T^2}\frac{\partial }{\partial F}$$

or

$$T\frac{\partial }{\partial T} = -\frac{1}{T}\frac{\partial }{\partial (1/T)}$$

Last edited: Apr 1, 2008
3. Apr 1, 2008

### knulp

Thanks, it makes complete sense now. So simple, doh!