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I'm having some difficulty understanding exactly what a partial derivative is.

I had been content with the definition

$$\frac{\partial F}{\partial x_i } = \lim_{\Delta x \rightarrow 0} \frac{F(x_0, x_2 ... x_{i-1}, x_i + \Delta x, x_{i+1} ... x_n ) - F(x_0 ... x_n )}{\Delta x} $$

However now I am beginning to realize that there is a bit more going on here than originally intended.

Consider in the definition of the Euler Lagrange Equations where given an operator

$$L (x,y(x), y'(x)) $$

We are seeking to find the optimal y for this operator. We are required to solve

$$\frac{\partial L}{\partial y} -\frac{d}{dx}[\frac{\partial L}{\partial y'}] = 0 $$

To find such L. But here's my issue. Where we treat all other variables constant except what we are deriving w.r.t,

How can you treat y constant and derive w.r.t. y' Because if either varies, then so does its complement.

For example consider expression $L = y^2$. If y = $e^x$ then

$$ \frac{\partial L}{\partial y'} = \frac{\partial e^{2x}}{\partial e^x} = 2e^x = 2y$$

But if $y = x^3$ then

$$ \frac{\partial L}{\partial y'} = \frac{\partial x^6}{\partial (3x^2)} = x^4 = y^{\frac{4}{3}}$$

In other words... what the hell? What does the partial derivative REALLY mean?

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    $\begingroup$ That's where using the $\partial_k F$ notation for the partial derivative with respect to the $k$-th argument helps avoid confusion. $\partial_2 L - \frac{d}{dx} \partial_3 L$ looks less confusing, doesn't it? $\endgroup$ – Daniel Fischer Dec 14 '14 at 22:09
  • $\begingroup$ I'm starting to understand, so a partial derivative doesn't actually take a function argument, ONLY an index $\endgroup$ – frogeyedpeas Dec 14 '14 at 22:12
  • $\begingroup$ But it also DOESNT care about any relationship two values share, where the values themselves are from different indices? $\endgroup$ – frogeyedpeas Dec 14 '14 at 22:12
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$y^{\prime}$ is an argument of $L$, so it makes perfect sense to (partial) differentiate with respect to it. Go through your question and mentally replace every instance of $y^{\prime}$ with $z$ and see if it makes more sense.

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  • $\begingroup$ That makes sense, at least the idea of, treat all other arguments constant, differentiate w.r.t. this one. But then the definition confuses me because other arguments also DEPEND on y' (namely y) so how to reconcile that? What are ALL the assumptions being made about partial derivatives $\endgroup$ – frogeyedpeas Dec 14 '14 at 22:14
  • $\begingroup$ $y$ and $y^{\prime}$ are just names in this context. $L$ is just a function of three variables. It so happens that when you give $L$ the inputs $x, y, y^{\prime}$, you get a function of $x$ which is a stationary point of a certain functional. But partial differentiating "with respect to $y^{\prime}$" does not "see" the fact that it's the ordinary derivative of $y$, because it's the same as differentiating $L(x,y,z)$ with respect to $z$ where $x,y$ can be any reals. $\endgroup$ – Nick Dec 14 '14 at 22:18
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    $\begingroup$ Just to hammer that point home, the fact that we're using $y$ and $y^{\prime}$ in stating the EL equation is completely arbitrary from the perspective of solving the equation. It's merely a linguistic imposition made to reflect the context in which the equation arises. $\endgroup$ – Nick Dec 14 '14 at 22:20

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