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I've been learning calculus for a week and I recently felt overwelmed about the amount of variables we use on derivatives and integrals. Can someone give me an overview about what the variables $x$, $d$, $dt$ and $h$ means? And if I've missed some very important ones for derivatives or intergrals please don't hesitate to point them out to me in the answers.

Thanks, Alex

AlexJaynMF
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    There is no fixed meaning for a particular variable; you can use any variables that you like. The commonly used ones are $x$ and $y$, at least in basic calculus. If you have any examples of specific variables being used that you are unclear about please share them and we can help clarify them. – A-Level Student Feb 01 '21 at 22:49
  • If you are getting confused about the differential $dx$, such a question has been asked before. See e.g. https://math.stackexchange.com/questions/1410774/what-allows-us-divide-multiply-dx-in-calculus – DanLewis3264 Feb 01 '21 at 22:50

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That's a pretty broad question, but here's a general answer:

  • You can usually think of $x$ as the "input" variable. Often you will see functions like $$f(x) = x^2+1.$$ Here $x$ is the input, and $f(x)$ is the output. The choice of "$x$" is just a convention, though; you could just as well write $$f(q) = q^2+1.$$ In either of these cases, the relationship is that "the output is equal to the input squared plus one".
  • $d$ by itself doesn't mean anything. If you see something like $dy/dx$, this indicates the (instantaneous) rate of change of $y$ with respect to $x$. This can be approximated by $\Delta y / \Delta x$, where $\Delta x$ and $\Delta y$ are the change, or difference (hence the letter $d$) in $x$ and $y$, respectively.
  • $dt$: this is the same as for $dx$ and $dy$ (see the previous note) but with respect to $t$ as opposed to $x$ or $y$.
  • $h$ is often used to represent a small increase in some variable. In that sense, you could just as well write it $\Delta x$.

Here's an example that combines some of these elements: if $y = x^2 + 1$, then $$\frac{dy}{dx} = \lim_{\Delta x \to 0}\frac{\left((x+\Delta x)^2+1\right) - (x^2+1)}{\Delta x},$$ or we could write exactly the same thing using $h$ instead of $\Delta x$: $$\frac{dy}{dx} = \lim_{h \to 0}\frac{\left((x+h)^2+1\right) - (x^2+1)}h$$

Théophile
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