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What is the motivation behind this definition? Could you give me an intuititive explanation?

We may also apply the differential notation to terms. If $\tau(x)$ is a term with the variable $x$, then $\tau(x)$ determines a function $f$.

$$\tau(x)=f(x),$$ and the differential $d\tau(x)$ has the meaning $$d\tau(x)=f'(x)dx\;.$$

Look at the bottom of page 58 and top of page 59 (and not 100-101) of https://www.math.wisc.edu/formMail/throttle.php?URL=/~keisler/chapter_2a.pdf

EDIT

As @Erick Wong so nicely adds in a comment: "The definition simply extends the meaning of $dy$ (where $y$ is a function of $x$) to an arbitrary expression that depends on $x$. ... But as in Example 5(a) it is natural to want to write $d(x^3) = 3x^2\, dx$ directly without first having to define a new function $y = x^3$ and then writing that as $dy$."

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    There are no steps. It is defined that way. See [Differential of a function](https://en.wikipedia.org/wiki/Differential_of_a_function) on Wikipedia. – 2017-02-26
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    @EvangelosBampas But I can almost see how you get there... there must be steps, the notation must have been derived in some way? I think it is just simple algebra, or something of the like... I may be wrong. I will look at your link, though – 2017-02-26
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    @AndreasAlmgren It's unclear what you mean. Why *must* there be steps to derive the meaning of a notation that has no meaning outside of the definition? What algebra do you imagine occurred between Example 3(a) and Example 5(a)? – 2017-02-26
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    @ErickWong I imagine 3a) as $y'=3x^2$, $y'=\frac{dy}{dx}$, $\frac{dy}{dx}=3x^2$, $dy=3x^2dx$. That makes me understand what I think happened, thus why $y=x^3$ "becomes" $dy=3x^2dx$. You do not agree:)? – 2017-02-26
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    @AndreasAlmgren The definition simply extends the meaning of $dy$ (where $y$ is a function of $x$) to an arbitrary expression that depends on $x$. Prior to this point in the book there was no such notation defined. But as in Example 5(a) it is natural to want to write $d(x^3) = 3x^2\, dx$ **directly** without first having to define a new function $y = x^3$ and then writing that as $dy$. – 2017-02-26
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    @ErickWong +1 on your comment. – 2017-02-26
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    @AndreasAlmgren I do believe you're asking the wrong question. Notation is not something that one **derives**. To derive something is to deduce it from more basic principles, but a definition is the most basic principle available, so this is an unreasonable expectation. It seems like you are more interested in the **motivation** behind the notation. That has more to do with the multiple tradeoffs one might consider when inventing notation: consistency, convenience, expressiveness, safety, naturalness. – 2017-02-26
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    @ErickWong What I'm askimg after is this: just as I show in 3(a) the steps before $dy=dx3x^2$, I want to know the algebraic steps between $\tau(x)=f(x)$ and $d\tau(x)=y'dx$. – 2017-02-26
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    @ErickWong And I am only 18 and not a mathematician, so I might not truly get why you are not liking my question. Not with intent of offending or the like, but with my level of ignorance "There are no steps. It is defined that way. See [Differential of a function](https://en.wikipedia.org/wiki/Differential_of_a_function) on Wikipedia." just seems idiotic. According to me, of course something that is defined has steps! Or how was it defined from the start, if not - and I know that may not be what the OP of the comment truly meant by writing "There are no steps". – 2017-02-26
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    If you insist on seeing a derivation of this, you have to define $df(x)$ *somehow* (since it's by no means evident what it means) and then we can talk about deriving $df(x)=f'(x)dx$ from that definition. I agree with @Erick that you probably intended to ask about the *motivation* behind the definition or, simply, about the intuition behind the notion of the differential of a function. – 2017-02-26
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    @EvangelosBampas Edited my question. – 2017-02-26
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    @ErickWong Edited my question. – 2017-02-26
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    @AndreasAlmgren I don't have time now but you should read from page 55 again to make sense, reasoning behind your equation is given from page 55 to 57. – 2017-02-27

2 Answers 2

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A somewhat shorter explanation: your equation says the same thing as $f'(x)=\frac{dy}{dx}$ or $dy=f'(x)dx$.

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Let $y= x^2\rightarrow (1)$

We assume that if $x$ is incremented by $dx$ then $y$ changes by $dy$ . If that's true then we have,

$y + dy = ( x+ dx)^2 \rightarrow (2).$

$$(2)-(1) = dy = 2xdx + (dx)^2$$

now $dx\rightarrow 0$ so $(dx)^2$ is even smaller hence $dy =2xdx$.

Similarly we go on for other functions. But its not possible to use this method to calculate differentials because when the functions are not polynomials or are composite then the method becomes difficult.

Just as we can use sum of series to calculate integrals. But its not possible to calculate the sum of every series that why we use the fundamental theorem of calculus. If you were asking this I hope it helps!