Why is the derivative defined as the limit of the quotient?

[fog37]

TL;DR

Understand derivatives as the limit of a quotient

Hello,
The derivative $\frac {dy} {dx}$ "appears" at first glance to be just the ratio of two infinitesimal quantities $dy$ and $dx$. However, infinitesimals are not really very very small numbers even if sometimes it is useful and practical to think about them as such. Infinitesimal are more like symbols than numbers since they don't important respect rules that numbers obey (Archimedean rule, etc.)
Numerically, a derivative is approximated and calculated as the ratio of two finite difference $\Delta y$ and $\Delta x$: $$\frac {\Delta y}{\Delta x}$$ For example, instantaneous speed is numerically calculated by forcing the time interval $\Delta t =t_2 - t_1$ to be finite and very very small. The space interval $\Delta x =x_2 - x_1$ does not have to be necessarily small: the object may undergo a large change in position over that short time interval hence having a large speed. So $\Delta x =x_2 - x_1$ does only need to be finite but not very very small at all.
$$v= \frac {\Delta x}{\Delta t}$$ No problem with that. I read that the actual theoretical derivative $\frac {dx}{dt}$ is NOT the quotient of the limit as $\Delta t$ and $\Delta x$ shrinks to zero but the limit of of the ratio $\frac {\Delta y}{\Delta x}$as $\Delta t \rightarrow 0$...

What is the difference between the concept of quotient of the limit of $\Delta x$ and $\Delta t$ versus the limit of the quotient ? Why is the derivative defined as the limit of the quotient?

Thanks!

 

[anuttarasammyak]

With t given
\frac{dx}{dt}=\lim_{\triangle t \rightarrow 0}\frac{x(t+\triangle t)-x(t)}{\triangle t}=\lim_{\triangle t \rightarrow 0}\frac{\triangle x}{\triangle t}
What's the problem on it ?

 

[Mark44]

No problem with that. I read that the actual theoretical derivative $\frac {dx}{dt}$ is NOT the quotient of the limit as $\Delta t$ and $\Delta x$ shrinks to zero but the limit of of the ratio $\frac {\Delta y}{\Delta x}$as $\Delta t \rightarrow 0$...

You have too many variables here: x, y, and t. Pick two.

What is the difference between the concept of quotient of the limit of $\Delta x$ and $\Delta t$ versus the limit of the quotient ? Why is the derivative defined as the limit of the quotient?

The first you wrote isn't meaningful and isn't the derivative. What you wrote is this"
$$\frac{\lim \Delta x}{\lim \Delta t}$$ presumably with both limits as $\Delta t$ approaches 0, and x being some function of t.

If x is a function of t, the derivative $\frac{dx}{dt}$ is defined as this limit quotient:
$\lim_{h \to 0}\frac{x(t + h) - x(t)}h$

 

[fog37]

You have too many variables here: x, y, and t. Pick two.
The first you wrote isn't meaningful and isn't the derivative. What you wrote is this"
$$\frac{\lim \Delta x}{\lim \Delta t}$$ presumably with both limits as $\Delta t$ approaches 0, and x being some function of t.

If x is a function of t, the derivative $\frac{dx}{dt}$ is defined as this limit quotient:
$\lim_{h \to 0}\frac{x(t + h) - x(t)}h$

Thanks,
The expression $$\frac{\lim \Delta x}{\lim \Delta t}$$ would seem to correspond to $$\frac{d x}{d t}$$
if we considered $dx$ and $dt$ the very small, infinitesimal, versions of $\Delta x$ and $\Delta t$...

What is the problem with that exactly? When we calculate the derivative numerically we are taking \Delta t## to be very small (not infinitely small)

 

[jbriggs444]

Thanks,
The expression $$\frac{\lim \Delta x}{\lim \Delta t}$$ would seem to correspond to $$\frac{d x}{d t}$$
if we considered $dx$ and $dt$ the very small, infinitesimal, versions of $\Delta x$ and $\Delta t$...

What is the problem with that exactly? When we calculate the derivative numerically we are taking \Delta t## to be very small (not infinitely small)

One might say that the problem is with the phrase "seems to correspond to". It is a similarity in form, not in substance. Limits do not commute in general.

$\lim_{\Delta x \to 0} \Delta x$ is zero, exactly.
$\lim_{\Delta t \to 0} \Delta t$ is zero, exactly.
$\frac{0}{0}$ is undefined.

 

[Stephen Tashi]

What is the difference between the concept of quotient of the limit of $\Delta x$ and $\Delta t$ versus the limit of the quotient ? Why is the derivative defined as the limit of the quotient?

"quotient of the limit" should be "quotient of the limits (plural)" since two limits are invoved.

To add to what @jbriggs444 has written, the notations
$lim_{x \rightarrow a} \frac{A(x)}{B(x)}$ and $\frac{ lim_{x \rightarrow a} A(x)} {lim_{x \rightarrow a} B(x)}$ obviously denote two different calculation procedures when we consider "taking a limit" to be a process. The first notation implies "taking a iimit" once and the second implies doing it twice.

So I'll assume your question is about why these different calculation procedures cannot be assumed to produce the same result. For example, why isn't there a theorem that says $lim_{x \rightarrow a} \frac{A(x)}{B(x)} = \frac{ lim_{x \rightarrow a} A(x)} {lim_{x \rightarrow a} B(x)}$?

If you search on the keywords "limit of a quotient", you will find that there is a such a theorem, but it has an exception. We must have $lim_{x \rightarrow a} B(x) \ne 0$. This exception isn't true in the type of limit used in taking derivatives. For a derivative, we have the situation $A(h) = f(t+h) - f(t)$ and $B(h) = (t+h) - h = h$ So $lim_{h \rightarrow 0} B(h) = 0$ in this case.

 

[fog37]

Thank you!

My confusion arises from reading $dx$ and $dt$ as both infinitesimally small quantities in the ratio $\frac {dx} {dt}$.

However $\frac {dx} {dt}$ derives from the limit as $\Delta t \rightarrow 0$. This does not mean (or does it?) that $\Delta x$ shrinks to zero too as $\Delta t$ does.

$\Delta x$ certainly gets smaller as $\Delta t \rightarrow 0$ but not necessarily infinitesimally small otherwise the ratio would become zero $\frac {dx} {dt}$. Clearly $\Delta t$ gets smaller faster than $\Delta x$ if the ratio needs to be nonzero.

I guess both $\Delta x$ and $\Delta t$ become smaller, almost zero, but at a different rate in this limiting process and never become actually zero. This "going to zero" for $\Delta t$ is just conceptual...

 

[Stephen Tashi]

I guess both $\Delta x$ and $\Delta t$ become smaller, almost zero, but at a different rate in this limiting process and never become actually zero.

That's an intuitive way of looking at things. However, if you read the formal definition of $lim_{x \rightarrow a} f(x) = L$, there is no mention of a process taking place in time or in steps. The idea that limit involves something "becoming" something at some "rate" is only an intuitive way of thinking about limits.

 

[anuttarasammyak]

I guess both Δx and Δt become smaller, almost zero, but at a different rate in this limiting process and never become actually zero. This "going to zero" for Δt is just conceptual...

You should remember x is function of t, x(t)
\triangle x = x(t+\triangle t)-x(t)
$\triangle x$ and $\triangle t$ are not independent small quantity.

 

[fog37]

You should remember x is function of t, x(t)
\triangle x = x(t+\triangle t)-x(t)
$\triangle x$ and $\triangle t$ are not independent small quantity.

Thanks anuttarasammyak.

Stephen Tashi is correct in the fact that I am trying to get an intuitive understanding of the derivative...that may be a mistake at the end of the day :)

I see how $x$ depends on $t$ and how $\Delta x$ and $\Delta t$ are not independent. That said, what should be the next step in my thinking? That as the quantity $\Delta t$ shrinks, also $\Delta x$ shrinks, correct?

 

[anuttarasammyak]

I recommend you to draw a xy graph, say $y=x^2$. Draw two points $A (x,x^2), B(x+\triangle x, (x+\triangle x)^2)$ on it and see how B changes as $\triangle x$ become smaller. How line AB
changes according to it ?

 

[fog37]

I recommend you to draw a xy graph, say $y=x^2$. Draw two points $A (x,x^2), B(x+\triangle x, (x+\triangle x)^2)$ on it and see how B changes as $\triangle x$ become smaller. How line AB
changes according to it ?

Hello,

I did. the line, when $\Delta x$ is large, is a secant. As $\Delta x$ gets smaller, tends to zero, the line becomes a tangent line. The $\Delta y$ and $\Delta x$ have different sizes. Both gets smaller as $\Delta x$ gets smaller faster...

 

[anuttarasammyak]

Now you observe value $\frac{\triangle y}{\triangle x}$ reaches slope of tangential line of the point you choose.

Say $y=x^2$, it is 2x where x is x coordinate of the point you choose.

 

[MidgetDwarf]

Ahh. I think you are referring to something called infinetismal calculus? I would first study calculus the "traditional" way as it is taught today. Then, if interested later, see the other approach.

My favorite calculus book is Moise: Calculus. It is a mixture of applied/theory which is closer to Courant Calculus, but not as difficult. Its well motivated, witty, and explanations are concise.

I also like the 3rd edition of Thomas: Calculus with Analytical Geometry. Do not get any other edition than the third. The proceeding editions are a completely different book. This book is also clear and concise, but it is an applied book.