# 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.

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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.

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You’re a braver man than I: I was going to wait with the $d/\Delta$ distinction until we got the big problem sorted! :-) –  Brian M. Scott Oct 5 '11 at 1:25
@Brian Yes, I am kind of torn between trying to be correct versus trying to give a useful answer. ;) Hopefully, the $d/\Delta$ distinction does not deter the OP from understanding what I say. Whether he uses it himself or not, it's up to his comfort level. –  Srivatsan Oct 5 '11 at 1:30
I know the difference between dy and $\Delta y$ but I guess I won't even ask about x. I am not too sure what is meant by $f(1.999) = y+\Delta y \approx f(2) + \Delta y.$ –  user138246 Oct 5 '11 at 1:32
@Jordan The second sign should also be equality. So $x$ changes slightly from 2 to $1.999$. What is the change in $y$? We calculated that before, and called it $\Delta y$. Now you know that the value of $f$ at $2$ is $16$. When $x$ changes now to $1.999$, it's new value (i.e., $f(1.999)$) is nothing but $f(2)$ plus whatever the change is. That's exactly what the equation is saying. –  Srivatsan Oct 5 '11 at 1:35
So how do I write that in a formula? $y= \delta y + y$? –  user138246 Oct 5 '11 at 1:37

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 $1.999^4$.

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y is 16 when x is 2. 15.968 Is this the proper way to do the problem? So y= y-dy? –  user138246 Oct 5 '11 at 1:27
@Jordan: Yes, apart from carefully distinguishing $d$’s from $\Delta$’s. That should be $y+dy$, though: you always add the differential in these problems. Finally, to avoid using $y$ for two different things at once, you should write $y\approx y_0+dy$. Here $y_0$ means the particular value of $y$ from which you’re starting; that’s $16$ in this case. –  Brian M. Scott Oct 5 '11 at 1:42
@Jordan: By the way, if you place the cursor on a formula, right-click, and click on Show Source, you can see the $\LaTeX$ code that was used to produce it; if you enclose that within dollar signs, you get the nice-looking displays. –  Brian M. Scott Oct 5 '11 at 1:47

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).

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Of course you're right in this example, but there are other cases where it's unnecessary and/or infeasibly complicated to do the exact calculation. –  leftaroundabout Oct 5 '11 at 10:10
Lol, yeah I know I'm just being a douche but figured some people might not realise it's actually pretty easy to be exact in this case :) –  user826788 Oct 5 '11 at 11:55

$\frac{d}{dx} x^4 = 4x^3$
$\frac{d}{dx} x^4$ at $x=2$ is $32$
So $1.999^5 \approx 2^4 - (32(0.001))$

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