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I've been using derivatives and integrals mechanically for years without really questioning the symbols. I recently watched some YouTube videos and came to understand that:$$\frac {dx}{dt}$$basically means, for some function, $f(t)=x$, an infinitesimal change in $t$, or $dt$, results in an infinitesimal change in $x$, or $dx$. The ratio of those two numbers is the derivative, or the instantaneous tangent line of $f(t)$ at $t$. So far, so good.

So could someone explain how to interpret this:$$\frac {d}{dt}$$I get that the bottom part is an infinitesimal change in $t$, but what is the top part? And how should I read an expression like $$\frac {d^2x}{dt^2}$$My main confusion is the $d$ part seems to have an existence on it's own without the dimension.

Quark Soup
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  • How do you interpret $'$ on it's own within the expression $f'(x)$? –  Oct 11 '19 at 02:51
  • I mean, $\frac d{dt}$ usually will be followed by a function $f(t)$, so you can interpret it as change in $f$ over change in $t$. Of course, the proper interpretation of derivatives relies heavily on exterior algebra and isn't particularly illuminating without further mathematical background, so this is a fine way of thinking of it. – Rushabh Mehta Oct 11 '19 at 02:51
  • You can interpret them by how they’re defined via limits: $\lim_{x\to c} \frac{f(x)-f(c)}{x-c}$. The functional viewpoint further allows us to define the operator, $\frac{d}{dt}$, by a mapping of one function $f$ to another function $f’$ defined by that above limit, whenever it exists. The geometric intuition is in there too—the difference quotient is the slope of the secant connecting $(x,f(x))$ to $(c,f(c))$, which tends to the slope of the (unique) tangent line at $c$ (if it exists) as $x \to c$, so also the interpretations as rates of change over infinitesimally small input changes, etc. – Nap D. Lover Oct 11 '19 at 03:03

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$\frac{d} {dt}$ is the differential operator that tells you to differentiate with respect to $t$. As such, $\frac{d^2x}{dt^2}$ means the second derivative of $x$ with respect to $t$.

For comparison, imagine seeing $\ln$ by itself or seeing $\tan$ by itself in the context of an equation -- it doesn't make any sense.

Andrew Chin
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    I wouldn‘t say $\log$ or $\tan$ don‘t make sense on their own - these are just functions. – Qi Zhu Oct 11 '19 at 02:55