I'm currently reading Thompson's Calculus Made Easy. Thompson states that given $y=x^2$ if we add $dx$ to x, and add $dy$ to y, then $y+dy=(x+dx)^2$.
My question though, is how can we assume that $y+dy=(x+dx)^2$? If I set $dy$ to 3, and $dx$ to 1, and set $y$ and $x$ to a constant, $y+dy$ would not be equal to $(x+dx)^2$ just because $y=x^2$.
So what about $dy$ and $dx$ ensures that $y+dy=(x+dx)^2$?
On
The key issue in differential calculus is "given a function $y=f(x)$, what happens to $y$ as $x$ moves a small amount?". To write that formally, we pick a small number which we'll call $\Delta x$. In practice, you could actually choose a value for this or better, a sequence of values decreasing to $0$. But whatever $\Delta x$ is, we then define $\Delta y$ as $$\Delta y = f(x+\Delta x)-f(x)$$ In words, $\Delta y$ is the change in $y$ for a given change in $x$.
Of course that relation can be rewritten as $$f(x)+\Delta y= f(x+\Delta x)\quad \text {or} \quad y+\Delta y = f(x+\Delta x)$$
Again, you may think of this as a purely numerical statement. If $f(x)$ is simple enough you could compute $\Delta y$ explicitly, given an explicit $\Delta x$.
This is what is meant in your text. They take it a little further in replacing the increments by differentials but I expect they aren't inclined to define the differential in any formal way...all that is intended is to consider the limit as $\Delta x $ tends to $0$.
Example. $y=x^2$
Then $$y+\Delta y = (x+\Delta x)^2=x^2+2x\Delta x+\Delta x^2\implies \Delta y = 2x\Delta x+\Delta x^2 $$
We arrange to get $$\frac {\Delta y}{\Delta x}=2x+\Delta x$$ and now we can pass to the limit as $\Delta x\to 0$ to get $$\frac {dy}{dx}=2x$$
So the numerical relations we have defined become the usual derivative in the limit.
$y$ is a function of $x$. What Thompson is trying to say is that when $x$ is increased by a tiny amount ($\mathrm{d}x$) then $y$ will also increase by a tiny amount ($\mathrm{d}y$). In other words, $\mathrm{d}y$ is the tiny change we cause in $y$ when we make the tiny change $\mathrm{d}x$.
Consider the figure below and assume that $\Delta x$ means the same thing as $\mathrm{d}x$. When we increase $x$ by $\mathrm{d}x$, we move from $A$ to $C$. Hence, the $x$ co-ordinate of $C$ is $x+\mathrm{d}x$.
But by moving to the right from $A$ to $C$, the function (the diagonal black line) has also moved from $A$ to $B$. But where has the function moved to when our $x$ co-ordinate is that of $B$ or $C$?
Since the function is $y=x^2$, we can see that at $B$, $y=(x+\mathrm{d}x)^2$. But our function has also moved up by $\mathrm{d}y$ so we say $y+\mathrm{d}y=(x+\mathrm{d}x)^2$.
In a nutshell, the $\mathrm{d}x$ is inside the brackets since we're performing the function on $x$. But the $\mathrm{d}y$ is outside the brackets since we're seeing how our $y$ changes when we make the change in $x$.
Does this make sense?
As pointed out in the comments, $\mathrm{d}x$ and $\mathrm{d}y$ aren't really specific numbers, as it's more useful to have them as algebraic variables. We're perfectly entitled to use specific numbers and say that $\underbrace{4}_y+\underbrace{0.42}_{\mathrm{d}y}=(\underbrace{2}_x+\underbrace{0.1}_{\mathrm{d}x})^2$ but keeping the terms as variables is instrumental in progressing to calculus.