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While i'm doing the math homework, I find something very strange. I am confused by a textbook's answer.

The Question

The textbook's answer

So, Does differentiation change the units of measurement in mathematical equations?

Nie Philip
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    It should be $cm^3s^{-1}$ – Andrei Aug 17 '23 at 15:45
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    The book is wrong. You and @Andrei are right. For the future: to post questions here don't use images. Write the math with mathjax: https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – Ethan Bolker Aug 17 '23 at 15:48

1 Answers1

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As @Andrei notes, the answer should be measured in $\mbox{cm}^3/\mbox{s}$. As far as the broader question is concerned, however, differentiation does indeed change the dimensions of physical quantities. Consider some function $f(x)$ depending on a physical quantity $x$. If you differentiate $f(x)$ with respect to $x$, the result will have the units of $f$, divided by the units of $x$; ie.

$$ \left[\frac{df(x)}{dx}\right] =\frac{[f(x)]}{[x]} $$

where $[\cdot]$ is shorthand for "the units of". If you have a time-dependent volume $V(t)$, then

$$ \left[\frac{dV(t)}{dt}\right] =\frac{[V(t)]}{[t]} $$

which reproduces the correct answer of $\mbox{cm}^3/\mbox{s}$, provided volume is measured in cubic centimeters and time is measured in seconds. My answer to this question may also be useful. Hope this helps!

CW279
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