# Leibnitz notation

1. Aug 17, 2005

### Severian596

Leibniz notation

Lately I've been having trouble with derivatives, specifically in the context of applying them to an ant problem (see this thread if you'd like to witness my slow decline). I have taken high school and college calc. I've learned many ways to perform the "hows" of calculus (how to find a derivative, how to find an integral, etc). But I admit it! I finally admit it to myself. I cannot apply calculus to complex problems.

Could someone briefly explain Leibnitz notation and how you can use it to better understand the meaning of derivative formulas? I already know how to take derivatives, so this is not really a concern. But when I first learned calculus they didn't stress the meaning(s) of notation enough, and now I feel left behind because I basically think of derivative notation as "f(x) with a prime symbol, like f'(x)."

I'll start slow. Can I define

(1) $$f(x) = x^2+3$$

then say that equations two and three below mean exactly the same thing?

(2) $$f'(x) = 2x$$
(3) $$\frac{d\ f(x)}{dx} = 2x$$

Last edited: Aug 17, 2005
2. Aug 17, 2005

### quasar987

Fo sho'.

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3. Aug 17, 2005

### Severian596

Great! Thanks Quasar. Now if I take equation (1) and rewrite it as

(4) $$y = x^2+3$$

to define y as a function of x, we have two variables: y and x. Therefore the differential of y with respect to x is

(4) $$\frac{dy}{dx} = 2x$$

Assuming that the above is true, my first question is this: what is the solution to (5), and what does it mean to you? (you know that y=x^2+3)

(5) $$\frac{dy}{dt} =$$

Essentially we differentiate by a variable that doesn't appear on the right side of this equation...asking, "How does a small change in time affect the value of y?" Correct? In notation that uses a prime only (remember f'(x)), it's assumed that you differentiated with respect to the variable x. But that's not always assumed using Leibniz notation, is it?

Last edited: Aug 17, 2005
4. Aug 17, 2005

### arildno

(3) is a notation that cannot really be misunderstood (hence, it is perfectly acceptable), but it is not wholly standard.
Given (1), defining f(x), a more conventional form of (3) would be:
$$\frac{df}{dx}=2x$$

Alternatively, neglecting to define "f", the following notation can also be used:
$$\frac{d}{dx}(x^{2}+3)=2x$$

In maths, you should strive for unambiguous and efficient notations; it is not necessary to slavishly follow those conventions you learn from your particular text-book if you don't like them.

5. Aug 17, 2005

### Severian596

Excellent! Thanks arildno...you know, the potential for someone to misunderstand this notation is high until someone says, "$d$ is an operator." I like the way you've given me three equalities, it drives home the point that $d$ is an operator.

6. Aug 17, 2005

### hypermorphism

If x is a function of time, for short if x=x(t), then we have by the chain rule (actually a theorem), dy/dt = (dy/dx)(dx/dt) = (2x)(dx/dt) where dx/dt is unknown because x(t) is unknown. If x is not a function of time, or x is a constant with respect to time, we have dx/dt=0, and thus dy/dt=0 by the above chain rule.
The meaning of the result is that if one were to assume y was a linear function of t at each t, then the resulting function is the change in y with respect to t (the slope) for each linear function at each t. Physical interpretations of this give things like velocity (change in position with respect to time), acceleration (change in velocity with respect to time), etc. Hence in your previous example in Homework, we said "assume that for small changes in t, the position x varies linearly with respect to t so that the change in position with respect to t is the sum of the velocities u and w", which we wrote as dx/dt = u + w for all x. You could also have written this as "assume dx = (u+w) dt" in analogy to the linear x = vt depending on how you think. Note that if w was a constant, the solution would have been the obvious x(t) = (u+w)*t, a straight line, representing a particle with constant velocity dx/dt = u+w.
The differential evaluated at each point is a way of reducing the study of nonlinear functions to the study of linear functions at each point, which we know much more intimately through linear algebra.

Last edited: Aug 17, 2005
7. Aug 17, 2005

### Severian596

I'd like to stress that I'm learning a lot today, including this. Essentially you didn't jump to the conclusion that the variable x was NOT a function. This is something that I'm not conditioned to do, but I'm learning to do this. I don't think schools stress the relationship between variable and functions enough...the fact that a variable can be a function if it's defined in terms of another variable is a logical leap that's easy once it's placed in front of you. Thank you hypermorphism! Back on track now...

Perfect, I see precisely. Excellent way of putting it. I just looked up the chain rule and a web page defined it as

$$\frac{d}{dx}f(u) = f'(u)\frac{du}{dx}$$

I wonder why they switched notations? It's easier to understand the notation of your equation, which would be

$$\frac{d}{dx}f(u) = \frac{df}{du}\frac{du}{dx}$$

8. Aug 17, 2005

### cscott

Have you gone through general differentiation using y = f(x)? I found it easier to see the leibnitz notation after seeing this process.

9. Aug 17, 2005

### HallsofIvy

Staff Emeritus
By the way, while a derivative $\frac{dy}{dx}$ is not a fraction (you don't find "dy" and "dx" separately and then divide!), it is the limit of a fraction and fraction "properties" can be proven by going back "before" the limit, using the "property", and then taking the limit again.

That's why you can think of the chain rule: $\frac{dy}{dt}= \frac{dy}{dx}\frac{dx}{dt}$ as "just cancel the "dx" terms" (but be careful not to say that out loud when your calculus teacher is listening!). It's not really correct but it is a good mnemonic.

Similarly, the rule for the derivative of an inverse function can be remembered as $\frac{dx}{dy}= \frac{1}{\frac{dy}{dx}}[/tex]. 10. Aug 18, 2005 ### Severian596 No I don't believe I've done this but I'm not quite sure what you mean, either. Sorry 11. Aug 18, 2005 ### cscott Do you understand how you end up with $$\frac{d}{dx}f(x) = \lim_{\Delta x \rightarrow 0} \left (\frac{f(x + \Delta x) - f(x)}{\Delta x} \right)$$ when you differentiate [itex]y = f(x)$? Sorry if I've confused you a bit, I just don't know what they go through in calculus classes as I've never had a proper one yet. :uhh: Are you asking where $\frac{dy}{dx}$ comes from?

Last edited: Aug 18, 2005
12. Aug 18, 2005

### Severian596

Well, constructing it in plain english maybe given (1):

(1) $$\frac{d}{dx}f(x)$$

you might ask, "How does one resolve the indicated operation on f(x) in (1)?" Or, "How does one find out how f(x) changes if x begins to approach zero?" I guess because that's the fundamental question behind every differential...what's the function's rate of change with respect to change with respect to some variable, right? So where does (1) lead us?

When trying to figure out rate of change you start with a snapshot of the function using some input value. Then you make a small change to the input value and find the difference between the new value of the function and the old value of the function. Finally you must divide the difference by the amount by which you changed the input, and you have a, "per incriment of change in the variable [fill in the blank]." That leaves you with

$$\frac{f(x + \Delta x) - f(x)}{\Delta x}$$

Is this what you mean? How am I doing?

13. Aug 18, 2005

### cscott

I think you have the idea but I'll add a bit more.

You first asked "Could someone briefly explain Leibnitz notation and how you can use it to better understand the meaning of derivative formulas?" so here it goes:

Consider any curve $y = f(x)$. Let $\Delta x$ be our increment to x and let $\Delta y$ be the corresponding increment to y.

We can now say

$$\Delta y = f(x + \Delta x) - f(x)$$

The gradient of the secante would be

$$\frac{\Delta y}{\Delta x} = \frac{f(x + \Delta x) - f(x)}{\Delta x}$$

Therefore the gradient at the point is

$$\lim_{\Delta x \rightarrow 0} \left ( \frac{\Delta y}{\Delta x} \right ) = \lim_{\Delta x \rightarrow 0} \left ( \frac{f(x + \Delta x) - f(x)}{\Delta x} \right )$$

$\Delta y$ becomes $dy$ and $\Delta x$ becomes $dx$.

$$\frac{dy}{dx} = \lim_{\Delta x \rightarrow 0} \left ( \frac{\Delta y}{\Delta x} \right )$$

or

$$\frac{d}{dx}f(x)= \lim_{\Delta x \rightarrow 0} \left ( \frac{f(x + \Delta x) - f(x)}{\Delta x} \right )$$

So my understanding is when we say $\frac{dy}{dx} =$ it represents this process

Last edited: Aug 19, 2005
14. Aug 19, 2005

### Hurkyl

Staff Emeritus
Not the process, just the end result of the process.

15. Aug 19, 2005

### Severian596

Thanks cscott, that illustration was very helpful. I'm becoming very comfortable with this notation, now. It feels good.