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In Calculus, the derivative w.r.t. $t$ of a single-variable function $f$ is denoted:

$$\frac{df}{dt}$$

While the derivative w.r.t. $t$ of a multi-variable function $f$ is denoted:

$$\frac{\partial f}{\partial t}$$

The justification being that the latter is distinct from the first in that it is a partial derivative.

Why does the latter warrant its own symbol? A single-variable function is just a special case of a function taking $n$ variables, so isn't the derivative of a single-variable function a partial derivative in a general sense?

The two operations seem to behave the same; is there any case where mixing them could be problematic?

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    $\begingroup$ If you want to discuss, for example, the derivative of $f$ in more than one variables along a curve $\gamma$, say, then $\frac{\text{d}f}{\text{d}x}$ and $\frac{\partial f}{\partial x}$ are totally different things. Note: $$\frac{\text{d}f}{\text{d}x}=\frac{\partial f}{\partial x}+\left(\frac{\partial f}{\partial y}\right)\left(\frac{\text{d}y}{\text{d}x}\right)+\text{similar terms for other variables}\,.$$ Only in one variable do the two notations coincide. $\endgroup$ – Batominovski Aug 20 '16 at 18:39
  • $\begingroup$ @Batominovski: Or you could write $\frac{∂(f ∘ γ)}{∂x} = \frac{∂f}{∂x} ∘ γ + (\frac{∂f}{∂y} ∘ γ)(\frac{∂y}{∂x})$ $\endgroup$ – user87690 Aug 23 '16 at 20:43
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This is taught very poorly in calculus courses, and you're confused because the notation is sloppy.

The insight you seek is the following1:

  • You take total derivatives of an expression with respect to a variable.

  • You take partial derivatives of a function with respect to its parameters.2

Before I go on, it's critical that you understand the following terminology:

  • A ("formal") parameter is a property of the function description itself.
    For example, the $a$ and $b$ in the function definition $f(a,b) = a + b$ are parameters.

  • An argument, or "actual" parameter, is a property of an expression that is a call to a function.
    For example, the $x$ and $y$ in the expression $g(x) + g(y)$ are arguments to $g$.

Now here's the kicker: if $h(x) = x^2$ then partial and total derivatives can be different:

\begin{align*} \frac{\partial}{\partial x} f(x, h)\ =\ 1\ \color{red}{\neq}\ 2x+1\ =\ \frac{d}{dx}f(x,h) \end{align*}

Makes sense? :-)

I hope it doesn't, because it was sloppy.

The notation above is extremely common, but not really correct.3

Remember I just told you partial derivatives are with respect to parameters whereas total derivatives are with respect to variables. This means that, if we've defined $$f(a,b) = a + b$$ as above, then it's actually incorrect (although very common) to write $$\frac{\partial}{\partial x}f(x,h)$$ for three reasons:

  1. $x$ is not a parameter to $f$, but an argument to it. The parameter is $a$.

  2. The second argument to $f$ should be a number (like $h(x)$), not a function like $h$.

  3. $f(x, h)$ is not a function, but a call to a function. It evaluates to a number.

So, to really write the above derivatives correctly, I should have written:

\begin{align*} \left.\frac{\partial f}{\partial a}\right|_{\substack{a=x\phantom{(h)}\\b=h(x)}}\ =\ \left.1\right|_{\substack{a=x\phantom{(h)}\\b=h(x)}}\ =\ 1\ \color{red}{\neq}\ 2x + 1\ =\ \frac{d}{d x} f(x, h(x)) \end{align*}

at which point it should be obvious the two aren't the same.

Makes sense? :)

1 This should be easier to understand if you know a statically typed programming language (like C# or Java).
2 You can define partial derivatives for expressions as well, but it'd just be implicitly assuming you have a function in terms of that variable, which you are differentiating, and then evaluating at a point whose value is also denoted by that variable.
3 Notice the expression wouldn't "type-check" in a statically typed programming language.

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In a single variable case, yes $\frac{df}{dx}=\frac{\partial f}{\partial x}$ but it's a bit odd to write this, because the whole point of a partial derivative is to emphasize that the $other$ variables are being treated as constants. Because you only deal with a single variable in Cal 1/2 (single variable calculus courses) it would seem pointless to mention partial derivatives. Also, the definition of the normal derivative is what you want in single variable calculus ,..... there will be times for both sorts of derivatives in multivariable calculus but only the normal derivative carries the same information that you learned from the single variable course.

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  • $\begingroup$ Nicely worded. +1 $\endgroup$ – Mark Viola Aug 20 '16 at 20:23

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