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When taking the derivative, such as $\frac{d}{dx}$, what exactly does the $d$ represent? The best answer so far is in for example $\frac{dy}{dx}$, the $d$ stands for change in and what follows the equality sign describes how $y$ changes in relation to $x$. But say the derivative is $3x^2 + y$, how exactly does that represent the relation between the two variables? I find it hard to wrap my head around.

rurouniwallace
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Chance
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  • The $d$ doesn't represent anything. You shouldn't look at $d,x$, and $\dfrac{\cdot }{\cdot}$ as individual symbols, but rather look at $\dfrac{\mathrm dy}{dx}$ as a whole symbol that just happens to be built from $d$s, $x$s and $\dfrac{\cdot}{\cdot}$s. And the symbol $\dfrac{\mathrm dy}{dx}$ represents the function $x_0\mapsto\lim \left(\dfrac{y(x)-y(x_0)}{x-x_0}\right)$ whenever the limit exists. Related question. – Git Gud Sep 05 '13 at 23:54
  • So if I shouldn't look at the d as a variable, how come I multiply by dx and dy when solving differential equations? – Chance Sep 06 '13 at 00:02
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    Because your lecturers choose to throw dirt in your eyes. That's not a multiplication at all, 'multiplying' is a cheap trick. The link in the comment above should help you with that. – Git Gud Sep 06 '13 at 00:04

2 Answers2

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The expression $\frac{dy}{dx}$ represents the limit of $\frac{\Delta y}{\Delta x}$ as $\Delta x$ vanishes. It's kind of an artifact of Leibniz's way of thinking about calculus in terms of infinitesimal numbers. Our modern expression $\frac{d}{dx}$ is an extrapolation from that, in which the $y$ has been pulled off to the side, where it can be replaced by whatever.

So, basically, the $d$ was originally an evanescent $\Delta$, and now it's just part of a notation.

G Tony Jacobs
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In $\dfrac{dy}{dx} = \dfrac{d}{dx}f(x)$, the quantity $dy$ or $df(x)$ is intuitively thought of as an infinitely small increment of $y=f(x)$ that corresponds to the infinitely small increment $dx$ in the variable $x$.

$dy=df(x)$ is measured in the same units as $y=f(x)$ (e.g. meters or seconds or dollars . . . ) and $dx$ is measured in the same units as $x$.

The second derivative is $\dfrac{d}{dx}\,\dfrac{d}{dx}\,y$, written as $\dfrac{d^2y}{dx^2}$.

The notation $dx^2$, then, means $(dx)^2$, not anything called $d(x^2)$ (which latter expression could be identified with $2x\,dx$). That implies that if $dx$ is measured in, for example, seconds, then $dx^2$ is measured in seconds squared. That is consistent with physical interpretations. If $dx$ is infinitely small, then $dx^2$ is of course infinitely small even by comparison to that already infinitely small quantity, so that would be consistent with thinking of $d^2y$ as a change that is infinitely small by comparison to the already infinitely small $dy$. But notice that $d^2y$ is measured in the same units as $y$, not in units of the square of that unit. That is also consistent with physical interpretations.

None of this is logically rigorous. The extent to which it can be made rigorous can bear, and has borne, a lot of examination in recent decades. That's a longer story.

Michael Hardy
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