# Prime notation for derivatives

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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It's not $\frac{\delta f}{\delta x}$, it's $\frac{df}{dx}$ (d, not delta). –  Arturo Magidin Dec 18 '10 at 7:13
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. –  Ｊ. Ｍ. 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.

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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. –  Cam Dec 18 '10 at 7:06
Personally, I wouldn't write $f=x^2$. I'd write $f(x)=x^2$ instead, but maybe this has already been discussed elsewhere. –  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, $\tfrac{df}{dx}(x^2+1) = f'(x^2+1)$.)

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