Differential to estimate a number I am trying to estimate $(1.999)^4$, so I set up the problem like this. $$y=x^4 ,$$ $$x=2 ,$$ $$dx=.001.$$ 
Then I find the derivative of $f(x)$ which is $4x^3$ and multiply that by $dx$ which is $.001$. This gives me an incorrect answer. 
 A: First, $dx = -0.001$, not $0.001$. Now $dy = 4x^3 dx$, so when you change $x$ by a small amount $dx$, you change $y$ by approximately $4x^3 dx$. In your problem $$4x^3 dx = 4(2^3)(-0.001) = -0.032;$$ this is the (approximate) amount by which you change $y$ when you move from $x=2$ to $x=2+dx=1.999$. 
$\qquad$(1) What is $y$ when $x=2$?
$\qquad$(2) If you change that $y$ value by $-0.032$, what do you get?
That is your approximation to $1.999^4$.
A: The notation you use is a little non-standard (in fact, technically wrong). I will use $\Delta x$ to denote what you call $dx$. 
Imagine the process like this. Suppose you are originally at $x=2$ and you move to $x+\Delta x = 1.999$. First of all, solving for $\Delta x$, we get $\Delta x = - 0.001$. The negative sign is crucial; it tells you whether the "new" value $x + \Delta x$ is larger or smaller than $2$.   (You missed the negative sign here.) 
Whatever you are calculating after that is the change in $y$ when the argument changes from $x=2$ to $x+\Delta x = 1.999$. This is given (approximately) by:
$$
\Delta y \approx f'(x) \Delta x .
$$
Here in place of $f'(x)$, you should plug in $f'(2) = 4 \cdot 2^3 = \ldots$. And, of course, $\Delta x = -0.001$. Multiplying these two numbers, you can find the value of $\Delta y$. 
But this is not what you were asked to calculate. This is the change in the value of the function as the argument changed from $2$ to $1.999$. You need to calculate $f(1.999)$. You can do so by:
$$
f(1.999) = y+\Delta y = f(2) + \Delta y.
$$
(You can understand this like: final value of the function = it's initial value + change in the function when the argument changed a little.) 
You know $f(2)$, and you know $\Delta y$. Can you calculate $f(1.999)$? 
Warning. You should be quite careful with the negative signs. I emphasize, again, that $\Delta x$ is negative. 
A: $\frac{d}{dx} x^4 = 4x^3$
$\frac{d}{dx} x^4$ at $x=2$ is $32$
So $1.999^5 \approx 2^4 - (32(0.001))$
A: Why not just get the exact answer?
$1.999^4 = (2-0.001)^4 = 2^4 - 4\times2^3\times0.001 + 6\times2^2\times0.001^2 - 4\times2\times0.001^3 + 0.001^4$
$= 16 - 0.032+0.000024-0.000000008+0.000000000001 = 15.968023992001$
Just thought I'd point out that you don't need a calculator. The $1, 4, 6, 4, 1$ of course come from Pascal's triangle. And yeah, $dx$ and $\Delta x$ both mean difference or change but one is infinitesimally small and more conceptual while the other is finite and definite (finite and de-finite... sounds like a contradiction).
