This may seem like an overly trivial question, but I've just recently become confused about Langrange's 'prime' notation for derivatives (for example $f'(x)$).

I know for sure that $f'(x) = \frac{\delta f(x)}{\delta x}$.

But suppose we replace x with an expression, like 2x+1. Do we write $f'(x^2+1) = \frac{\delta f(x^2+1)}{\delta x}$ or $f'(x^2+1) = \frac{\delta f(x^2+1)}{\delta (x^2+1)}$?

Does putting the prime around the function instead of between its letter and parentheses make a difference? For example what does $(f(x^2+1))'$ mean?

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    $\begingroup$ It's not $\frac{\delta f}{\delta x}$, it's $\frac{df}{dx}$ (d, not delta). $\endgroup$ – Arturo Magidin Dec 18 '10 at 7:13
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    $\begingroup$ Cam, if you're going for compactness of notation, but the limited utility of primes is troubling you, one compact notation I've seen used a capital D: $D_x \sin\;x=\cos\;x$ for instance. $\endgroup$ – J. M. is a poor mathematician Dec 18 '10 at 10:02

$f'$ is a function, so $f'(2x + 1)$ denotes $f'$ applied to $2x + 1$, or $\frac{df}{dx}(2x + 1)$. For example if $f = x^2$ then $f' = 2x$ and $f'(2x + 1) = 4x + 2$.

$(f(x^2 + 1))'$ is the derivative of the function $f(x^2 + 1)$, which is $2x f'(x^2 + 1)$ by the chain rule.

This question highlights a weakness of the $'$ notation, which is that it always comes with an implied variable with respect to which you're differentiating. If this variable is clear from context there's no problem, but sometimes it isn't.

  • $\begingroup$ Thanks! I wanted to be able to use the prime notation because it's shorter than $\frac{\delta f(x)}{\delta x}$, but I found it too ambiguous at times, so this is great. $\endgroup$ – Cam Dec 18 '10 at 7:06
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    $\begingroup$ Personally, I wouldn't write $f=x^2$. I'd write $f(x)=x^2$ instead, but maybe this has already been discussed elsewhere. $\endgroup$ – Hendrik Vogt Dec 18 '10 at 9:09

Often people write something like $(x^2)'=2x$ or $(e^x)'=e^x$, but as you noticed yourself, this is ambiguous, and I never use this notation. You can write $\frac{d}{dx}x^2=2x$ instead, i.e., you don't have to put the $x^2$ into the numerator. This way it looks much clearer, I think (though not shorter, unfortunately). So for me it would be $$ \tfrac{d}{dx}f(x^2+1) = 2xf'(x^2+1). $$ (However, $\tfrac{df}{dx}(x^2+1) = f'(x^2+1)$.)


I am reading Quiaochu Yan when answering Cam:

1) $𝑓′$ is a function, so $𝑓′(2𝑥+1)$ denotes $𝑓′$ applied to $2𝑥+1$, or $\frac{\partial 𝑓}{\partial 𝑥}(2𝑥+1)$. For example if $𝑓=𝑥^2$ then $𝑓′=2𝑥$ and $𝑓′(2𝑥+1)=4𝑥+2$

2) $(𝑓(𝑥^2+1))′$ is the derivative of the function $𝑓(𝑥^2+1)$, which is $2𝑥𝑓′(𝑥^2+1)$ by the chain rule

and I agree with Quiaochu but I don't understand why such difference is a weakeness of the $′$ notation. We see that $𝑓′(2𝑥+1)$ and $(𝑓(𝑥^2+1))′$ are written differently, and they mean different things. There is no ambiguity, I think. I am more puzzled about Hendrik Vogt's post, when he says that $(𝑥^2)′=2𝑥$ is ambiguous, where is the ambiguity? On lessons, under the constrains of time, I use that notation constantly on introductory Calculus courses, wherever and whenever there is only one variable with respect to which we are differentiating with. I noticed that such notation is more common in European countries than in the UK and rare in the United States, to the extent that some colleagues told me that it was wrong to write $(𝑓(𝑥^2+1))′$.


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