# I Delta x in the derivation of Lagrange equation

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1. Nov 24, 2017

### jamalkoiyess

Hello PF,

I was doing the derivation of the Lagrange equation of motion and had to do some calculus of variations.

The first step in the derivation is to multiply the integral of ƒ(y(x),y'(x);x)dx from x1 to x2 by δ.

and then by the chain rule we proceed. But I cannot understand why we are not applying the chain rule to x too. All the proofs omit this part saying that x1 x2 are fixed and thus it goes to zero.

The book I am using is "Classical dynamics of Particles and Systems" by Marion and Thornton.

Thank you!

2. Nov 24, 2017

### Orodruin

Staff Emeritus
It is not a multiplication by delta. It is the variation of the integral based on variations in the functional argument y(x).

3. Nov 24, 2017

### Delta²

x is independent from the variations of y(x) as a function (but x depends on the variations of the values of the function y(x)).

What we mean by "variations of y(x) as a function" are other functions $g_{\epsilon}(x)=y(x)+\epsilon\eta (x)$ for some fixed function $\eta(x)$ and for any $\epsilon>0$. So what we basically mean is that x is independent of $\epsilon$.