23
$\begingroup$

When presented with an equation, say, $y=5x^3+7x^2+4x+9$, you can write on the second line, $\frac{dy}{dx}=15x^2+14x+4$. Similarly, $f(x)=5x^3+7x^2+4x+9$ and $f'(x)=15x^2+14x+4$. But is there a way to write "the derivative of $5x^3+7x^2+4x+9$ is $15x^2+14x+4$" in just one line?

What should l write, $\frac{dy}{d5x^3+7x^2+4x+9}=\cdots$? That fraction just gives me a headache trying to understand it.

What about $f'(5x^3+7x^2+4x+9)=\cdots$? For all the reader knows, $f(x)$ could be anything, and the writer wanted them to plug in $5x^3+7x^2+4x+9$ into the original $f(x)$ and then take the derivative.

So has anyone come up with a better way to write this that does not involve defining anything and then using the newly defined function/operator?

$\endgroup$
9
  • 54
    $\begingroup$ "$f'(x^3 + 5x) = 3x^2 + 5$" is grossly wrong and is done by zillions of confused freshmen. If, for example, $f(x) = \sin x$, then $f'(x^3 + 5x) = \cos(x^3 + 5x)$. This is not the same as plugging in $x^3+5x$ and then taking the derivative: that would yield $\cos(x^3+5x)\cdot(3x+5).$ I once saw an assertion on an exam that $f'(5)=0$ because $f'$ is a derivative and $5$ is a constant. That is very thoroughly missing the point. $\endgroup$
    – Michael Hardy
    May 16, 2017 at 20:05
  • 2
    $\begingroup$ \dots sometimes appears as \cdots and sometimes as \ldots depending on the context, thus: $a + \dots + b$ looks like $a+\cdots+b$ but $a,\dots,b$ looks like $a,\ldots,b$. In $a+b+c+\cdots$ and $a=b=c=\cdots,$ \cdots seems appropriate because that's how \dots would be rendered if any binary relation or binary operation symbol followed. $\endgroup$
    – Michael Hardy
    May 16, 2017 at 20:15
  • 7
    $\begingroup$ As for the mathematical question, the short answer is: because of the chain rule. If you write $f'(5)$ that means plugging $5$ into the function $f'$; it does not mean plugging $5$ into $f$ and then differentiating. The latter would yield $0$. And if you understand what $g(5)$ means, where $g$ is a function, and that $g'$ is a function, then the above follows logically and trivially. $\endgroup$
    – Michael Hardy
    May 16, 2017 at 20:15
  • 7
    $\begingroup$ Regarding your first sentence, are you suggesting that $\frac{d}{dx} = 15x^2+14x+4$ in your example? If so, that is incorrect. The derivative of $y$, which is $\frac{dy}{dx}$ is equal to that, not the operator $\frac{d}{dx}$ itself. $\endgroup$
    – zahbaz
    May 16, 2017 at 23:32
  • 2
    $\begingroup$ When you say "Should I write it as"..., you're almost right! You just replaced the wrong variable. You're taking the derivative of the polynomial with respect to $x$, not the derivative of $y$ with respect to the polynomial. (The polynomial is $y$.) Most of the time, people just "take it out of the fraction" as if it were being multiplied (even though it isn't) to make it look nicer. $\endgroup$
    – Deusovi
    May 17, 2017 at 17:10

5 Answers 5

Reset to default
127
$\begingroup$

You would denote the derivative of $5x^3+7x^2+4x+9$ as $$\frac{d}{dx}(5x^3+7x^2+4x+9)$$ That is the only notation I've ever seen unless the expression is expressed as a function.

$\endgroup$
14
  • 33
    $\begingroup$ $D_{x}(5x^3 + \cdots)$ is also common. See en.wikipedia.org/wiki/Differential_operator#Notations $\endgroup$
    – BlueRaja - Danny Pflughoeft
    May 17, 2017 at 7:29
  • 20
    $\begingroup$ Yet another notation (which I sometimes use) is $(5 x^3 + 7 x^2 + 4 x + 9)'$ when it is clear that which variable we are differentiating with respect to. $\endgroup$
    – Alex Vong
    May 17, 2017 at 12:28
  • 2
    $\begingroup$ @origimbo: Unless I'm missing something, Joker_vD seems to be using bare parentheses as a symbol for differentiation. Which famous mathematician proposed that? $\endgroup$
    – ruakh
    May 18, 2017 at 1:37
  • 12
    $\begingroup$ @AlexVong: No, while there is only one way to make sense of $(5 x^3 + 7 x^2 + 4 x + 9)'$, it is in fact confusing two notations: the prime is an operation on functions where $\frac d{dx}$ is a transformation of expressions. It is true that the way mathematics is taught there is lots of implicit conversion form expression to function and back, but what you write does that twice which ends up being confusing; made explicit it means $(x\mapsto 5 x^3 + 7 x^2 + 4 x + 9)'(x)$. Differentiation $\frac d{dx}$ is what we learn to do at school, and then $(x\mapsto E)'=(x\mapsto\frac{dE}{dx})$. $\endgroup$
    – Marc van Leeuwen
    May 18, 2017 at 9:59
  • 2
    $\begingroup$ @PaoloLeonetti Sigh... yeah, I hate it when this kind of thing happens. I get $2$ or $3$ votes on some of my best answers, but $126$ on a short answer about notation. Isn't life ironic? :P $\endgroup$
    – Franklin Pezzuti Dyer
    Jun 28, 2017 at 16:53
45
$\begingroup$

A common choice of notation is $D_{x}(5x^3 + 7x^2 + 4x + 9)$. The subscript indicates the variable with respect to which one is differentiating.

$\endgroup$
2
  • $\begingroup$ If there's only one variable, then subscript $x$ is not needed. I usually write something like $D \{5x^3+7x^2+4x+9\} = 15x^2+14x+4$. $\endgroup$
    – md2perpe
    May 19, 2017 at 11:52
  • 3
    $\begingroup$ I have also used the notation $5x^3+7x^2+4x+9 \overset{d/dx}{\longrightarrow} 15x^2+14x+4$. $\endgroup$
    – md2perpe
    May 19, 2017 at 11:55
34
$\begingroup$

The most common choice is $\frac{d}{dx}$. If the variable is clear from context, you can use a plain $D$.

If you have several variables and you only want to differentiate with respect to one, it's best to write it as a partial derivative with $\frac{\partial}{\partial x}$ or $\partial_x$.

I have also seen notations like $(5x^3+7x^2+4x+9)'$ or $(5x^3+7x^2+4x+9)_x$, but I would strongly recommend using $\frac{d}{dx}$ instead.

There are several kinds of derivatives, and it's good to use notation that is compatible with them (uses similar syntax). It is easy to replace $\frac{d}{dx}$ with a $\frac{\partial}{\partial x}$, a $\nabla$, a $\Delta$ or a $d$.

$\endgroup$
2
  • 7
    $\begingroup$ Some people prefer a more upright "d", as in $\frac{\mathrm{d}}{\mathrm{d}x}$ instead of $\frac{d}{dx}$, but that is just typography. I suppose nobody is so sick as to use $d$ as a variable with which to differentiate: $\frac{\mathrm{d}}{\mathrm{d}d}(5d^3)$ $\endgroup$
    – Jeppe Stig Nielsen
    May 18, 2017 at 9:33
  • $\begingroup$ @JeppeStigNielsen I was typographically lazy. The $d$ is indeed often upright, but the convention is not uniform. Any sort of D is a bad variable name unless there is a strong reason to use it. $\endgroup$
    – Joonas Ilmavirta
    May 18, 2017 at 15:07
5
$\begingroup$

Yes, there is one. As $f'$ represents the derivative of $f$, you can use the prime symbol like this:

$$(5x^3+7x^2+4x+9)' = 15x^2+14x+4$$

I have already seen it being used like that. Also note that as long as you don't make it confusing for the reader, you can make up your own notation if it's useful.

$\endgroup$
4
$\begingroup$

Just as the symbol $$\int(\cdots)\;dx$$ denotes the antiderivative of something (the expression where the "$\cdots$" is), so the symbol $$\frac{d}{dx}(\cdots)$$ denotes the derivative of something (again, the expression where the "$\cdots$" is).

For example, you would have $$\frac{d}{dx}(13x^2-27x+1) = 26x-27$$ just as you would have $$\int(26x - 27)\;dx = 13x^2-27x + C$$

Occasionally "$D$" or "$D_x$" is seen in lieu of "$\frac{d}{dx}$", and it is very frequent to use an appended "prime" or "apostrophe" to mean the same thing, as $$(13x^2-27x+1)' = 26x-27$$

In other words, $$(\cdots)'\equiv \frac{d}{dx}(\cdots)$$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .