1. The problem statement, all variables and given/known data http://www.mathpages.com/home/kmath523/kmath523.htm In trying to arrive at the Lagrange equations of motion, on the above website, I stumbled over a bit that involved reversing the order of differentiation in the equation (4): d(partial)/d(partial)X1[(dx/dt)] = d/dt[d(partial)x/d(partial)X1] the way i wrote it may be confusing. you can check the website for the original equation. 2. Relevant equations The author justifies this reversal with d/dt = d(partial)/d(partial)t and the commutative property of partial differentiation. Quote:"Since x and X are both strictly functions of t, it follows that partial differentiation with respect to t is the same as total differentiation, and so the order of differentiation in the right-most term of (4) can be reversed (because partial differentiation is commutative)." I buy that taking the partial derivative of x w/t is the same as d/dt, since here x= f(t), so I feel fine with writing d(partial)/d(partial)t instead of d/dt when it is on the outside. but if i then move the partial derivative w/t to the inside using the commutative property of partial derivatives, I cannot show that this is justified. 3. The attempt at a solution Since taking the derivative of x w/respect to t, results in a function of t only, if i then take its partial derivative w/respect to X1, I get always zero? But if i take the partial derivative first w/respect to X1, i get a function of X1 that I can take the derivative w/respect to t, and get a function of t? I get two different results depending on the order I take derivatives! How can this be justified? x=f(X1,Y1) for example, x= X1 +Y1 = t^2 + 2t d(partial)/d(X1) [x] = 1 d/dt(2t) = 0 I arrive at 0 d/dt[x] = 2t+2 d(partial)/d(X1) [2t+2] = 0? or 2(X1)^1/2 ---> = X1^-1/2 ? I get always zero or something different... what about x = X1^2 + Y1 = t^4 +2t = 2X1 =4t and =4t^3+2 =0? or =4t? since 4t^3+2 = 4t(X1) +2 Ok. I think i see an explanation. If x is first order in X1, then result is zero. -That was the first example. If x is higher order than 1 in X1, the resulting expression in t contains X1 so therefore can be rewritten as f(t,X1) giving the desired result of f(t) when partial differentiated w/respect to X1 But rewrite t in terms of X1 *only* if they are exactly equal. t does not equal (t^2)^1/2, so does not depend on t^2. taking the partial derivative w/respect to t^2 then yields nothing this explains why the first result is zero and not (X1)^-1/2 So the final explanation is that taking d/dt of x does not destroy any X1 that is left over, so I'm free to partial differentiate in X1 afterwards. If i take partial derivative in X1 first, i only get no X1 left if i have first order, which would result in d/dt(c) = 0 anyways. If its higher order, then I get a function of X1 which differentiates nicely w/respect to t. The partial derivative in X1 from the first method can be rewritten in t, as X1=f(t), so in the end we both get a nice function of t. I see how it works now, but i haven't proved it. Any corrections or further explanations are welcome! if you can show me the proof I'd be glad to read it, thanks.