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Is it correct to say 'dee squared y over dee x squared'?

I am just trying to make sure because I have to teach this to someone.

I learnt that $f''(x)$ is pronounced 'f double prime of x' but I couldn't find how to pronounce Leibniz notation. Thanks!

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    $\begingroup$ I don't think there is any "official" way to read this symbol. I would just call it the second derivative. $\endgroup$ – user9464 Jan 18 at 0:49
  • $\begingroup$ It's the second derivative indeed $\endgroup$ – imranfat Jan 18 at 0:49
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    $\begingroup$ The way you pronounce it is fine. That's what I say. $\endgroup$ – Lee Mosher Jan 18 at 0:52
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    $\begingroup$ I think people often leave out the word "over" $\endgroup$ – J. W. Tanner Jan 18 at 1:12
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    $\begingroup$ I always say "second derivative" but in my heart I say "dee two y, dee x squared". Have you checked Khan Academy? $\endgroup$ – Neat Math Jan 18 at 3:04
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Three ways of reading $\dfrac {d^2 y}{dx^2}$...

  1. "dee squared y over dee x squared" (your way)
  2. "dee two y over dee x two" (informal)
  3. The second derivative of $y$ with respect to $x$ (formal)
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    $\begingroup$ Did you mean the second derivative of $\color{red}y\,?$ $\endgroup$ – J. W. Tanner Jan 18 at 1:33
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    $\begingroup$ Saying “squared” between “dee” and “ex” resolves the issue addressed with the asterisk $\endgroup$ – gen-ℤ ready to perish Jan 18 at 1:42
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    $\begingroup$ Your (1) and (2) are missing a word like "over" or "by" corresponding to the horizontal bar in the fraction. $\endgroup$ – Rob Arthan Jan 18 at 2:05
  • $\begingroup$ @RobArthan: isn't that word commonly elided? $\endgroup$ – J. W. Tanner Jan 18 at 2:21
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    $\begingroup$ "dee two y over dee x two" is the best! much more natural. $\endgroup$ – Bertrand Wittgenstein's Ghost Jan 18 at 3:43
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Formally, you should say : second order derivative of y w.r.t. x

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I'm pretty sure you read it either

  • dee two y by dee x squared

or

  • dee two y over dee x squared

(either is fine)

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You could read as: "dee two y over dee x two"

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  • $\begingroup$ No! $R2D2 \neq R^2D^2$ (put $R =D =1$ to see why not). $\ddot{\smile}$ $\endgroup$ – Rob Arthan Jan 20 at 0:55
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Yes, we read it this way. But it is actually the second derivative of the function $y$ which is a function of $x$.

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