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I was watching one of the Khan Academy videos on differential equations ("Exact Equations Intuition 1 (proofy)") and there's something that confused me.

In the video, they use both the derivative of a function $\psi(x, y(x))$ with respect to $x$,

$$\frac{d}{dx}\psi(x, y(x)),$$

as well as the partial derivative of the same function with respect to the same variable,

$$\frac{\partial \psi}{\partial x}$$

(I think) I understand what a partial derivative of a function is (you consider its other arguments constants and you essentially turn it into a derivative of a single variable function), but I don't understand what a non-partial derivative with respect to one variable means. How is it different from a partial derivative?

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You may be overcomplicating this. (I don't mean that as an insult, I mean that helpfully.) There isn't any significant difference between a "non partial derivative of a function" and a "derivative of a function". (In this case. If we were being cute, we would say, "A non partial derivative is any derivative which is not a non partial derivative. (i.e. It could be a classic derivative, a directional derivative, a total derivative, etc. etc.)) I presume you have learned what a derivative is, so what you're looking at here is just a derivative of $\psi$ with respect to $x$. – 000 Jan 16 '12 at 21:05
up vote 7 down vote accepted

For the partial derivative $\frac{\partial \psi}{\partial x}$, you consider $\psi$ as a function of two independent variables $x$ and $y$, and see how that changes when you vary $x$ while holding $y$ constant. For $\frac{d\psi}{dx}$, you are letting $y$ be a function of $x$ (that's what the $y(x)$ means), so when $x$ changes $y$ also changes. The bivariate version of the chain rule says

$$ \frac{d}{dx} \psi(x,y(x)) = \frac{\partial \psi}{\partial x}(x,y(x)) + \frac{dy}{dx}(x) \frac{\partial \psi}{\partial y}(x,y(x))$$

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I would use the "evaluated at" bar notation instead, since this was confusing to me when I read your answer. – Mehrdad Apr 4 at 18:51

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