# Partial Derivatives versus Proper Derivatives

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?

• 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? – Daniel Fischer Dec 14 '14 at 22:09
• I'm starting to understand, so a partial derivative doesn't actually take a function argument, ONLY an index – frogeyedpeas Dec 14 '14 at 22:12
• But it also DOESNT care about any relationship two values share, where the values themselves are from different indices? – frogeyedpeas Dec 14 '14 at 22:12

$y^{\prime}$ 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.
• $y$ 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. – Nick Dec 14 '14 at 22:18
• 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. – Nick Dec 14 '14 at 22:20