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What’s the difference between the notation $(\sin(x))^\prime$ and $\sin^\prime (x)$? Where prime denote the derivative of some function. This question seems trivial at first. If we change the argument of function to something different expression of x. The outcome becomes completely different. To give an example, consider the function $\sin(\frac{1}{x})$. We have $(\sin(\frac{1}{x}))^\prime=\sin^\prime(\frac{1}{x}) (\frac{1}{x})^\prime = - \frac{1}{x^2} \cos(\frac{1}{x})$. In this example, we treat $\sin$ function as composite function. We are usually custom to write $(f(x))^\prime$ as $f^\prime (x)$. Perhaps we can not apply this “notation” to this given function. Although, I don’t known the precise definition of $\sin$ function, I am assuming (due to lack of my knowledge) that, it is defined on $\mathbb{R}$, and it’s range is $[-1,1]$. Since, $(\sin(x))^\prime$ and $\sin^\prime (x)$ are equal for $x \in \mathbb{R}$. Isn’t $\frac{1}{x} \in \mathbb{R}$ is also a real number? So, why do we have different outcome for derivative operation? Is $\sin$ function is really a composite function?

Warning: My question generally contains error and/or misunderstanding of concept. If you find something wrong then mention it on comment.

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  • $\begingroup$ As a general rule, one should strive to avoid ambiguous notation...though I would have said that both of these mean$\frac d{dx}\sin (x)=\cos(x)$. Note, too, that $\frac d{dx}\frac 1x=-\frac 1{x^2}$, you lost a sign somewhere. $\endgroup$
    – lulu
    Commented Nov 12, 2021 at 18:32
  • $\begingroup$ I suppose that they mean the same, I would recommend to avoid using $(\sin(x))'$ and stick with $\sin'(x)$ when you mean "the derivative of the function sine evaluated at the point $x$" as some people (myself included) may interpret $(\sin(x))'$ as "the derivative of the constant function whose value equals $\sin(x)$" (of course, such derivative is zero and then I would wonder why you wrote such a convoluted expression for zero!). $\endgroup$
    – William M.
    Commented Nov 12, 2021 at 18:35
  • $\begingroup$ @lulu yeah. I’m sorry for my error. I will fix it. $\endgroup$
    – user264745
    Commented Nov 12, 2021 at 18:39
  • $\begingroup$ @WillM. Thank you for telling the different interpretation of $(sin(x))^\prime$ notation. $\endgroup$
    – user264745
    Commented Nov 12, 2021 at 18:48
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    $\begingroup$ Long story short: I would always stick with $\sin'(x).$ A similar story for $\sin(x)$ over $\sin x.$ $\endgroup$
    – William M.
    Commented Nov 12, 2021 at 19:23

2 Answers 2

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Technically, $[\sin(x)]'$ is an abuse of notation, since $\sin(x)$ is not itself the function being differentiated, but the output of the function being differentiated: it is an element in the range of the function. So, in any case, if you want to embrace the pedantry, you should use the notation $\sin'(x)$ instead, because now it is clear that you are evaluating the function $\sin'$ at $x$ to obtain the output $\sin'(x)$.

However, there is another reason why you should keep in mind a distinction between $[\sin(x)]'$ and $\sin'(x)$. Suppose we naively replace $x$ with $f(x)$ instead. Now, one would typically interpret $[\sin(f(x))]'$ as an expression where the chain rule is applied: hence you would simplify this as $\cos[f(x)]f'(x)$. But you would never interpret $\sin'(f(x))$ as $\cos[f(x)]f'(x)$. Instead, you would interpret it as $\cos[f(x)]$, since you are taking $\sin'=\cos$, and you are taking $f(x)$ as the input. Because of these shenanigans, you can actually write the equation $$(\sin[f(x)])'=\sin'[f(x)]f'(x),$$ and this would be technically correct, even though it definitely does not look correct. So the fact is that a distinction is intended to exist, but it works in a way that is extremely misleading. This is the other reason why you should avoid using the notation $[f(x)]'$ and stick to $f'(x)$ instead. And if you really need a notation that allows you to use the chain rule, then instead of writing $(\sin[f(x)])',$ you should write $(\sin\circ{f})'(x)$ to avoid confusion. This makes the distinction between $\sin'[f(x)]$ and $(\sin\circ{f})'(x)$ clear and without any abuse of notation. Hence you can write $$(\sin\circ{f})'(x)=\sin'[f(x)]f'(x),$$ and now the chain rule as applied makes sense visually.

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You correctly note that the use of the prime to indicate differentiation is ambiguous in this context. I've never seen it and you should not use it. If you actually encounter it somewhere, hope the context makes the author's meaning clear.

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  • $\begingroup$ I think, it is a pretty standard notation used in Baby Rudin book. $\endgroup$
    – user264745
    Commented Nov 12, 2021 at 18:37
  • $\begingroup$ To be more specific, example 5.6. $\endgroup$
    – user264745
    Commented Nov 12, 2021 at 18:41
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    $\begingroup$ @user264745 I'm surprised. I hope it's the unambiguous $\sin^\prime$ that's found there. $\endgroup$ Commented Nov 12, 2021 at 18:42
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    $\begingroup$ @user264745 The best notation for derivatives depends on the context. $df/dx$, $dy/dx$, $f^\prime$, $y^\prime$, $Df$ all have their place. I would never write $f(x)^\prime$. If you need the variable name and the prime the prime goes on the function name. $\endgroup$ Commented Nov 12, 2021 at 19:06
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    $\begingroup$ @user264745 ...see, consider again the example with $\sin(1/x)$. It is clear that since $\sin'=\cos$, you would write $\cos(1/x)$. But this is not the same as doing $[\sin(1/x)]'=\cos(1/x)(-1/x^2)$, applying the chain rule. It turns out that $[\sin(1/x)]'$ is the wrong notation here, but in order to be able to denote the expression correctly, you need to change the way you denote the reciprocal function. The issue is that the expression $1/x$ is not itself a function, but rather, the output of a function. What you should do is... (see next comment) $\endgroup$
    – Angel
    Commented Nov 12, 2021 at 22:57

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