Differentiating $y=x^{2}$

Question

I am reading in a book about differentiating, but I am confused with one of the steps he takes. We start with:

$$\begin{align} y &= x^{2} \\ y + \mathrm{d}y &= (x + \mathrm{d}x)^2 \\ y + \mathrm{d}y &= x^2 + x\mathrm{d}x + x\mathrm{d}x + (\mathrm{d}x^2) \end{align}$$

Now the author simplifies this to:

$$y + dy = x^2 + 2x\mathrm{d}x + (\mathrm{d}x^2)$$

I dislike how the middle term is simplified to $2x\mathrm{d}x$ instead of $2(x\mathrm{d}x)$, as I feel like it is more intuitive on what is going. As in, $2$ of the term $x\mathrm{d}x$, instead of $2x\mathrm{d}x$. But I fear writing it as $2(x\mathrm{d}x)$ may result in an incorrect distributive property.

Next, he omits the $(\mathrm{d}x^2)$: $y + \mathrm{d}y = x^2 + 2x \mathrm{d}x$.

Subtract the original $y = x^2.$:

$$\mathrm{d}y = 2x \mathrm{d}x.$$

Now here is where I get confused:

$$\frac{\mathrm{d}y}{\mathrm{d}x} = 2x.$$

How can he just divide both sides by $\mathrm{d}x$!? If the original term was $2$ of $x\mathrm{d}x$, wouldn't it have to be written out as $2x * 2\mathrm{d}x$, and thus divide both sides by $2\mathrm{d}x$ instead?

I think the root of my confusion is how to properly simplify:

$x\mathrm{d}x + x\mathrm{d}x$.

I trust that he is right, but I am looking for an explanation of why his simplification can work, and why $2(x\mathrm{d}x)$ would be incorrect.

Answer 1

Your question is a good example of what happens when people work with infinitesimals outside of non-standard analysis: a lot of confusion. Look, the notion of $dx$ being a very tiny $x$ (less than any real number, and yet nonzero) is not precise, and it's not possible to define it correctly in standard analysis. There are some people that try defining $dx$ as $\Delta x$ when $\Delta x$ goes to zero, but this is zero by the definition of limit, so this is just garbage.

Many people ask: "why should we care if it is rigorous or not?", and well, it's just because when working with something rigorous the chance of confusion is much less than with something that is not even defined.

In the rigorous framework, we let $f: \Bbb R \to \Bbb R$ be given by $f(x)=x^2$, then by the definition of the derivative we have:

$$f'(a)=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}=\lim_{h\to0}\frac{(a+h)^2-a^2}{h}=\lim_{h\to 0}\frac{a^2+2ah+h^2-a^2}{h}$$

Now we simply reorganize the last expression, getting:

$$f'(a)=\lim_{h\to 0}\frac{(2a+h)h}{h}=\lim_{h\to0}2a+h=2a$$

So this limit exists for every $a \in \Bbb R$ and thus $f$ is differentiable with derivative $f'(x)=2x$ at every $x \in \Bbb R$.

So my suggestion is that you abandon this "intuitive" notion of infinitesimals and move to the rigorous standard analysis. You can pick Spivak's Calculus book: it's a very good book, even for self-study, and it'll show you how to deal with all of these things in a rigorous and straightforward way.

Answer 2

Multiplication is associative, just as is addition:

That is, with addition, we know that $a + (b + c) = (a + b) + c = a + b + c$.

(In other words, parentheses can be omitted without causing any ambiguity).

The same is true with multiplication:

$$2\cdot (a\cdot b) = (2\cdot a)\cdot b = 2\cdot a \cdot b$$

and with multiplication, we often simply "juxtapose" the terms, omitting "$\cdot$" or "$\times$" to get $2 \cdot a \cdot b = 2ab$.

Now in your question, you are asking about simplifying: $$x\mbox{d}x + x\mbox{d}x\tag{1}$$

Here we can use the distributive property of multiplication over addition:

$$ab + ab = (a + a)b = (2a)b = 2ab$$

So, applying this to $(1)$: $$x\mbox{d}x + x\mbox{d}x = (x + x)\mbox{d}x = (2x)\mbox{d}x = 2x\mbox{d}x\tag{2}$$

Answer 3

$$2(ab) \neq 2a\cdot 2b = 4ab$$

$$2(ab) = 2a(b) = a(2b)$$

Either you take the 2 with the $x$, or with the $dx$.

[The above is assumed to be over reals]

Answer 4

"If the original term was $2$ of $x \, dx$, wouldn't it have to be written out as $2x\cdot 2 \,dx$, and thus divide both sides by $2 \, dx$ instead?"

Don't let the infinitesimal stuff throw you for a loop. I know the pure math guys are going to tear their hair out when I say this, but at your stage of the game you can just treat dx like a number. Specifically, where you have $2(xdx)$, substitute 1 for $x$ and $0.1$ for $dx$. Then you have $2(1*0.1)$. Work it out. Does this give you $2\times1\times2\times0.1=0.4$? You probably see that it doesn't. It gives you $2\times 1\times 0.1=0.2$. In the same way, $2(xdx)=2\times x \times dx = 2xdx$.

My answer is basically the same as amWhy's earlier answer, but I thought a concrete example would be helpful.