Why does replacing dx and dy with finite values yield this result? 
*

*Consider the change in the function $f(x)=x^2$ at $x=1$  when $x$ is changed by some value $\Delta x$.
$$
\Delta y = f(x+\Delta x)-f(x)\\
\to \Delta y = (1+\Delta x)^2-1^2
$$
Let's set this value $\Delta x$ to $0.1$. Then
$$
\Delta y = (1+0.1)^2-1^2=0.21
$$

*Now let's consider this change in the function utilizing its derivative.
$$
\frac{dy}{dx}=2x\\
dy=2x(dx)
$$
Let's replace the infinitesimal differences $dx$ and $dy$ with a finite difference (is this valid?).
$$
\Delta y = 2x(\Delta x)
$$
Again, let's set $\Delta x$ to 0.1.
$$
\Delta y = 2x(0.1)\\
\to\Delta y=2(1)(0.1) = 0.2
$$
Why does $\Delta y$ end up being different than it was previously? I would think that since we're replacing $dx$ with a finite difference, $dy$ would be the same as  $\Delta y$ in #1. There must be a flaw in my thinking somewhere.


Edit:
Thanks for the answers. Here's the justification I worked out.
$$
\frac{\Delta y}{\Delta x} = \frac{(x + \Delta x)^2 - x^2}{\Delta x}\\
\frac{\Delta y}{\Delta x}=\frac{x^2+2x\Delta x + (\Delta x)^2 - x^2}{\Delta x}\\
\frac{\Delta y}{\Delta x}= 2x+\Delta x\\
\to\Delta y = \Delta x(2x + \Delta x)\\
$$
As $\frac{\Delta y}{\Delta x}$ approximates $\frac{dy}{dx}$ with smaller and smaller values of $\Delta x$, a remainder term is always present. Since $x^2$ is exponential, $\Delta y$ must be greater than multiplying the current growth rate, $f'(x)$ by $\Delta x$.
Using 0.1 for $\Delta x$ gives us
$$
\Delta y = 0.1(2(1) + 0.1) = 0.21 
$$
the same value calculated in #1.
 A: By definition of derivative of a function as $f'(a)=\lim\limits_{h\to 0}\frac{f(a+h)-f(a)}{h}$, it follows that we can write
\begin{align}
f(a+h)-f(a)&= f'(a)h + R(h)
\end{align}
whereby the "remainder term" $R(h)$ is "small" in the sense that it satisfies $\lim\limits_{h\to 0}\frac{R(h)}{h}=0$.
If you want to modify the notation, we can write this as
\begin{align}
\Delta f_a(h) &= df_a(h) + R(h)
\end{align}
As you can see, this equation shows in general, the actual difference $\Delta f_a(h)$ is NOT equal to the linear approximation $df_a(h)$ (afterall that's why we call it an approximation), because of the extra term $R(h)$.

Or if you use the classical Leibniz notation which suppresses all the arguments, you can write this as $\Delta y = dy + R$, for some "small" remainder term, $R$. So as you can see, it is NOT true that $\Delta y = dy$. The correct way of writing things is that
\begin{align}
\Delta y &= \frac{dy}{dx}\cdot \Delta x + R
\end{align}
or simply $\Delta y \approx \frac{dy}{dx}\cdot \Delta x$. This is only an approximation, NOT always an equality. It is only an equality for affine functions (i.e of the form $y=mx+b$).
A: Note that $\displaystyle \frac {dy} {dx} = \lim_{\Delta x \to 0} \frac{\Delta y} {\Delta x} $.
That limit is very important. Taking $\frac {dy} {dx} \approx \frac{\Delta y} {\Delta x} $ is an approximation that holds only for small values of $\frac{\Delta x} {x} $, and even then, only approximately (although the approximation gets better and better as $\frac{\Delta x} {x} \to 0 $).
So your first algebraically calculated value was exact, while the second value using calculus was an estimate, albeit a reasonably good one, since $\frac{\Delta x} {x} = \frac{0.1}{1} = 0.1 $, a fairly small value.
A: The derivative of a function is not the value of
$$
\frac{\Delta y}{\Delta x}
$$
for any finite value of $\Delta x$, but rather the limit of the above ratio as $\Delta x \to 0$. More formally, if $y=f(x)$, then
$$
f'(x)=\lim_{\Delta x \to 0}\frac{\Delta y}{\Delta x}=\lim_{\Delta x \to 0}\frac{f(x+\Delta x)-f(x)}{\Delta x} \, .
$$
If you pick a very small value of $\Delta x$, such as $\Delta x=0.1$ as you did in your question, then you will find that
$$
\frac{\Delta y}{\Delta x} \approx \frac{dy}{dx} \, .
$$
Thus, the derivative can be informally considered as the change in $y$ divided by the change in $x$ for small changes in $x$, even though, as we have discovered, this is not strictly the case.
