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Let's consider a following equation: $$x''+x'+x=5$$ My question is: how should I differentiate it with respect to x'? I've read somewhere that you only differentiate x' and take both x and x'' as constants, but why? Intuition tells me that they somehow depend on x'.

EDIT: I need this to compute a lagrange equation, where you take the derivative of L with respect to q'.

Lugi
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  • I thinkl you should read $x''(t)+x'(t)+x(t)=5$. This is a second order differential equation and the unknown is the function $x(t)$. – Claude Leibovici Apr 18 '15 at 12:41
  • Velocity often doesn't depend on position. E.g. if velocity = 4km/hr, you don't need to know the position in space to work out the velocity. But in general it depends on the nature of x(t) whether it is dependent on position or not. E.g. acceleration due to gravity is dependent on position, and thus you would have to apply the chain rule when differentiating. – Kenshin Apr 18 '15 at 12:43
  • But how should I differentiate it in this particular case? – Lugi Apr 18 '15 at 12:46
  • Are you trying to find the Euler-Lagrange Equation? Or a first integral? – Matthew Cassell Apr 18 '15 at 13:23
  • You can find a nice answer to your question here. Physicists like to talk about the derivative with respect to x', but what is really ment is the partial derivative of L with respect to the third component. – abc Apr 18 '15 at 13:24

3 Answers3

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Note that, under suitable conditions, the chain rule applies with $\frac {df}{dt}=\frac {df}{dy}\cdot \frac {dy}{dt}$

Use this with $f=x, x', x''$ and $y=x'$.

Mark Bennet
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  • Note the addition of the comment about the Lagrange equation makes other answers rather more relevant. I'm leaving this as "how to, if you need to" – Mark Bennet Apr 18 '15 at 20:24
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For that you would need an actual Lagrange functional. Something like $$ L[x]=\int_a^b\left(\frac12 \dot x(t)^2-V(x(t))\right)\,dt $$ Then you apply it to a perturbed curve $x+δx$, or more precisely to a family $x+s·δx$ of perturbed curves, and compute the part linear in $δx$ $$ δL[x;δx]=\lim_{s\to0}\frac{L[x+δx]-L[x]}s = \int_a^b\left(\dot x(t)·δ\dot x(t)-V'(x(t))·δx(t)\right)\,dt $$ Then apply partial integration to get rid of the derivative of $δx$ and set all coefficients of $δx(t)$ to zero.

Only to summarize the form of the so-obtained Euler-Lagrange equations do you use derivatives by $\dot x$ as if it were an independent variable.

Lutz Lehmann
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Let $v = \frac{dx}{dt}$, then $$\frac{d^2x}{dt^2} = \frac{dv}{dt} = \frac{dx}{dt}\frac{dv}{dx} = \frac{v}{\frac{dx}{dv}}$$

Equation becomes $$ \frac{v}{\frac{dx}{dv}} + v + x = 5 $$

Differentiating with respect to $v$ $$\frac{1}{\frac{dx}{dv}} - \frac{v}{\left(\frac{dx}{dv} \right)^2}\frac{d^2x}{dv^2} + 1 + \frac{dx}{dv} = 0$$

$$ \left(\frac{dx}{dv}\right)^3 + \left(\frac{dx}{dv}\right)^2 + \frac{dx}{dv} - v\frac{d^2x}{dv^2} = 0 $$

Dylan
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