A Question On Differentiation On $\frac {dy}{dx}$

Question

I'm learning how to differentiate a parabola, and to establish that $\frac{dy}{dx} = 2x$.

The textbook I'm using is trying to verify that $\frac{dy}{dx} = 2x$. It states "suppose $x=100$, and therefore $y=10,000$. Then let $x$ grow till it becomes $101$ (that is $dx=1$). Then the enlarged $y=101^2=10,201$. But if we agree that we may ignore small quantities of the second order, $1$ may be rejected as compared with $10,000$; so we may round of the enlarged y to $10,2000$, therefore $$\frac{dy}{dx}=200/1=200.$"

My question is that why can you round off the $1$ in $10,201$ to make $10,200$? In what situations is it acceptable to do this? And if you can regularly do this, then how can calculus be accurate? Or is calculus just an approximation?

Answer 1

I suggest looking at more values: $$ (100 + 1)^2 = 100^2 + 2 \cdot 100 \cdot 1 + 1^2 = 10201 \\ \Longrightarrow \frac{(100 + 1)^2 - 100^2}{1} = 201 $$ $$ (100 + 0.1)^2 = 100^2 + 2 \cdot 100 \cdot 0.1 + 0.1^2 = 10020.01 \\ \Longrightarrow \frac{(100 + 0.1)^2 - 100^2}{0.1} = 200.1 $$ $$ (100 + 0.01)^2 = 100^2 + 2 \cdot 100 \cdot 0.01 + 0.01^2 = 10002.0001 \\ \Longrightarrow \frac{(100 + 0.01)^2 - 100^2}{0.01} = 200.01 $$ The less we grow $x$ the less does the second order term influence the difference ratio; it gets closer to $200$. The argument in the book is just meant to very unrigorously exemplify this fact.

Answer 2

If your text book really says that, then I agree with Hans Lundmark- throw away that text book and get a better one! You cannot do derivatives without some kind of limit process. Yes, the derivative of $y= x^2$, at x= 100, is 200. But not because "we may ignore small quantities of the second order". Taking x= 100 and "h" to be any x value, not just "1", then y(100+ h)= (100+ h)^2= 10000+ 200h+ h^2 so the increase from x= 100 to x= 101 is y(100+ h)- y(100)= 10000+ 200h+ h^2- 10000= 200h+ h^2 (in the special case that h= 1 that is 201) and the "difference quotient", the increase in the y value divided by the increase in the x value, so the average rate of change from x= 100 to x= 100+ y, is $\frac{200h+ h^2}{h}= 200+ h$. Now, we do NOT just "ignore" that last h- we take the limit as h goes to 0 to get the rate of change at x= 100 to be 200.

The "limit" concept is essential to the derivative.

Answer 3

You are asking:

In what situations is it acceptable to round off the $\mathbf{1}$ in $10,201$ to make $10,200$?

in the context of differentiation.

Two conditions must hold together:

$1)$ When, as in this case, there is a positive number (in this case $101 - 100 = 1$) such that for all $x$ that are far away from $100$ not much than this positive number (that is, such that $\vert dx\vert$ is less than this positive number), $y$ takes values $x^2$ that rounds off to $10,000+2x\,dx$ with an error less than $\mathbf{1}$

For instance for $x=100.1$ that is distant apart from $100$ less than what $101$ is from $100$, it results that $100^2+2\cdot 100\cdot 0.1$ is a better approximation of $y=100.1^2$ than your original $100^2+2\cdot100\cdot1$ was with respect to $101^2$. That is, the error is less than your original $\mathbf{1}$.

$2)$ Where the approximation in the sense of $1)$ can be done for any other positive number (not only $\mathbf{1}$).

For instance, take $\mathbf{0.1}$. There are correspondingly positive numbers (one of which is $0.1$, for instance) such that for $dx$ in magnitude not greater than it, it results that the round off error due to the approximation of the correct value of $y(x+dx)=(x+dx)^2$ with the estimated one $y(x)+dy=x^2+2x\,dx$ satisfies $1)$.

Then you ask:

is calculus just an approximation?

Not at all. The differentiation is not a way for correct values of the increment of a function to be substituted for with its linear approximation, which would result in what you now think calculus is, but is a way to associate to a function (like, for instance, $y=x^2$) a "higher-level function" (a map) that to every $x$ of the original function associates a new function: the linear approximation of the increment of the original function in that same point. Such a linear approximation is called the differential of the function at the given point, while the higher-level function is simply called the differential of the function.

In your case, for instance, at the point $100$ it is associated the differential of $y$ at $100$ that is given by $2\cdot100\,dx$, while the whole differential of $y$ is $dy=2x\,dx$.

So from the original function you got a family of linear functions in $dx$ indexed by $x$ (that is, not only a single linear approximated function).

With this complex entity, many problems can be solved, among which that of reconstructing the original function, given the value of it in a single point only, by integration. Indeed you can look at it the other way around: have a look at how the integration is defined in order to appreciate why it is appropriate that differentials are defined the way they are for them to solve the reconstruction problem by integration.

The mathematical analysis of differential functions is highly developed thanks to the linearity of the differential (with respect to $dx$) that allows the use of the well understood theory of linear algebra.