# Splitting the derivative?

1. May 26, 2012

### KingBigness

1. The problem statement, all variables and given/known data

From
$\frac{de_{s}}{dT} = \frac{L_{v}e_{s}}{R_{v}T^{2}}$

derive

$e_{s}(T) = 6.11 e^{\frac{L}{RV}(\frac{1}{T}-\frac{1}{273})}$

2. Relevant equations

3. The attempt at a solution

The way my lecturer derived it was he 'split' the derivative and took them to their respective sides and integrated.

So he got

$\frac{de_{s}}{e_{s}} = \frac{LdT}{R_{v}T^{2}}$

However I was under the impression that you can't 'split' a derivative like that. Is this just a shortcut some physicists take to make the maths more simple? If it is what is the correct way of deriving this?

2. May 26, 2012

### Curious3141

3. May 26, 2012

### KingBigness

How accepted is it? I am doing a double degree in pure mathematics and physics. If I was to do this in my pure maths classes would I be stoned?

4. May 26, 2012

### iRaid

I've taken calculus 1 and I've seen it before? :s

5. May 26, 2012

### KingBigness

Sorry I wasn't very clear. I have seen it before and done it countless times I just always have a memory of a teacher saying don't tell a pure mathematician about it.

6. May 26, 2012

### SammyS

Staff Emeritus
In my experience, physicists & engineers are notorious for treating Leibniz's notation for the derivative as if it were a fraction.

A more rigorous handling of this derivative equation might be as follows:
Treating es as a function of T we have

$\displaystyle \frac{d}{dT}(e_{s}) = \frac{L_{v}e_{s}}{R_{v}T^{2}}$

Rewriting this equation gives us

$\displaystyle \frac{1}{e_{s}}\frac{d}{dT}(e_{s}) = \frac{L}{R_{v}T^{2}}$

Integrating w.r.t. T gives

$\displaystyle \int {\frac{1}{e_{s}}\frac{d}{dT}(e_{s})}dT = \int{\frac{L}{R_{v}T^{2}}}dT$

We can rewrite the integral on the left hand side.

$\displaystyle \int {\frac{1}{e_{s}}\frac{d}{dT}(e_{s})}dT = \int {\frac{1}{e_{s}}}\,de_{s}$​

Alternatively, if $\displaystyle \frac{d}{dT}(e_{s}) = \frac{L_{v}e_{s}}{R_{v}T^{2}}\,,$
then the differential of es is given by $\displaystyle d\,e_{s} = \frac{L_{v}e_{s}}{R_{v}T^{2}}\,dT$

7. May 27, 2012

### KingBigness

exactly what I wanted. thank you