# Homework Help: Don't understand d squared

1. Jan 5, 2007

### Dr Game

This is a rather basic question, I'm just learning... but I don't understand the difference between

d^2y / dx^2

and

dy^2 / dx^2

I have included an image where I drew it if you don't understand. I can see how they aren't the same, but its more really I don't get what d^2y means...

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2. Jan 5, 2007

### d_leet

I'm not sure if it's entirely a matter of notation or not, I hope someone else will be better able to answer that for you, but the first image represents the second derivative of y with respect to x, and the second image represents the square of the derivative of y with respect to x.

3. Jan 5, 2007

### gabee

I'm interested in this too, because I've never learned the exact reason for the notation. However, now that I think about it, I think this could be an explanation.

The differential operator is

$$\frac{d}{dx}$$

which means to take the derivative of something with respect to x. Used, it looks like this:

$$\frac{d}{dx}y = \frac{dy}{dx}$$

To take the second derivative of something, you would differentiate it twice:

$$\frac{d}{dx} \frac{d}{dx} y = \bigg( \frac{d}{dx}\bigg)^2 y = \frac{d^2 y}{dx^2}$$

Like d_leet said, $$\bigg( \frac{dy}{dx} \bigg)^2$$ is the square of the derivative of y with respect to x.

I hope this is right and if anyone can give a clearer explanation I would be glad to hear it.

Last edited: Jan 5, 2007
4. Jan 5, 2007

### Dr Game

I can understand that.

Do you know what d squared means? Its just is that, thats something new to me.

alright.. lets say was y=x^4 how are the two different

Last edited: Jan 5, 2007
5. Jan 5, 2007

### gabee

The way I understand it, d sort of means "an infinitesimal amount of." So $$\frac{dy}{dx}$$ means "an infinitesimal amount of y divided by an infinitesimal amount of x," which is the same thing as $$\lim_{\Delta x \to \infty} \frac{f(x+\Delta x) - f(x)}{\Delta x}$$, the definition of a derivative. Doing it twice leads to the "squared" notation.

Last edited: Jan 5, 2007
6. Jan 5, 2007

### d_leet

What two are you referring to here? If you mean the two expressions in your original post then the first evaluates as 12x2, and the second evaluates as 16x6 for y=x4.

7. Jan 5, 2007

### Dr Game

I'm going to fail, I can't get that...

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8. Jan 5, 2007

### d_leet

I can't make any sense of what you did. d is not a number or a function, in this sense(the sense you used it in, I suppose) it isn't really anything at all, it is part of an operator as in
$\frac{d}{dx}$ the differential operator, but I have no clue as to what you tried to do, or why in the world you have +C attached to your final "answer." It looks like you don't really understand differentiation, and are even possibly confusing it with integration by adding C to your final answer.

9. Jan 5, 2007

### Hurkyl

Staff Emeritus
Well, what you wrote was gibberish.

$$\frac{d}{dx}$$

only has two parts to it:

$$\frac{d}{d\ }$$

and

x.

The d's don't have an independent existence: they are merely part of the one big symbol d/d_. There is no division going on there.

Writing

$$\frac{d}{dx} \frac{d}{dx}$$

as

$$\frac{d^2}{dx^2}$$

is a notational shortcut suggested by the appearance of the symbols. It is not meant to be literally squaring the d.

Last edited: Jan 5, 2007
10. Jan 5, 2007

### gabee

Dr Game, it looks like you tried to solve for d somehow, but you have to remember that d is not a variable, it's part of an operator, like + or - or $$\sqrt{\;}$$. You can't solve for the value of an addition sign! It's the same way with the differentiation operator; it performs an action on what comes after it.

Last edited: Jan 5, 2007
11. Jan 5, 2007

### d_leet

d by itself, however, as Hurkyl pointed out doesn't really have a meaning, it is part of an operator, but it, alone, isn't the operator.

12. Jan 5, 2007

### gabee

Ah, right, I probably should say $$\frac{d}{dx}$$ is the operator.

Last edited: Jan 5, 2007
13. Jan 5, 2007

### Dr Game

$$\frac{d^2 y}{dx^2}$$

$$\frac{d^2 x^4}{dx^2}$$

$$\frac{x^4}{x^2}$$ <-- d's cancel...

14. Jan 5, 2007

### d_leet

We've said this several times already that $$\frac{d^2}{dx^2}$$ is not a fraction, it's not a function, a number or anything of that sort, it is an operator that takes the second derivative of whatever it is operating on. In the case above you should be taking the second derivative of x4 not simplifying something that is not a fraction in the first place.

15. Jan 5, 2007

### Hurkyl

Staff Emeritus
What I'm about to say is completely unrelated to the subject of this thread, for the reasons we've said.

The notation ab² means a*b², not (a*b)².

16. Jan 5, 2007

### Dr Game

well if all you do is take the second derivitive, I've solved and gotten $$12x^2$$ but it doesn't make that much sense to me... I think I need to try and get more into this.. I am just new to DE

17. Jan 5, 2007

### d_leet

What doesn't make sense to you? And I want to point out that aside from notation, and the notion of taking derivatives, very little in this thread has anything to do with differential equations.

18. Jan 5, 2007

### Dr Game

I'm going to go with the classic exuse that I've had a lot of really bad calculas teachers in the past and never leant what an 'operator' is... I just read the section in my text book, and this opened up everything, I see things clearly now...

you derive it because it is an operator...

$$\frac{d^2 y}{dx^2}$$

$$\frac{d^2 x^4}{dx^2} = 4x^3$$

$$\frac{d 4x^3}{dx} = 12x^2$$

then the other one is simply...

$$\frac{dy^2}{dx^2}$$

$$\frac{d^2 (x^4)^2}{dx^2}$$
is this right so far.. because when you take the derivitive of this.. you get $$8x^7$$ then second derivitive as $$56x^6$$ which isn't right...

19. Jan 5, 2007

### gabee

You got the first part right (in the final result), but I'm guessing the second part you wanted to do this:

$$\bigg(\frac{dy}{dx}\bigg)^2 = \bigg( \frac{d}{dx} x^4 \bigg)^2 = ( 4x^3 )^2 = 16x^5$$

ALWAYS read your textbook, especially if you don't understand something. :) If you still have trouble it then you should ask your teacher personally. The textbook will sometimes explain things that the teacher did not or that you didn't pick up on in class.

Last edited: Jan 5, 2007
20. Jan 5, 2007

### Dr Game

why do you only take the derivative once... is not everything else squared too? you have to derive it twice don't you to get rid of $$\frac{d}{dx}$$

21. Jan 5, 2007

### gabee

Mmm...no, because the $$dy \over dx$$ has to be evaluated immediately. Then you square it. $$\bigg( \frac{dy}{dx} \bigg)^2$$ does NOT equal $${d^2 \over dx^2} y^2$$.

Last edited: Jan 6, 2007
22. Jan 5, 2007

### d_leet

I honestly don't think you can blame your misconceptions on a bad calculus teacher, but that is really beside the point so I won't make any more comments about that. Again, being honest when I first learned differential calculus my teacher never spoke of $$\frac{d}{dx}$$ as an operator, I picked that up later, but again that isn't particularly relavent to your confusions.

Ok you took the first derivative here not the second, what you should have on the right hand side should be $$\frac{d}{dx}[4x^3]$$.

This part is correct.

As written this is correct, notationally $$\frac{dy^2}{dx^2}$$ represents you taking the second derivative of y2 which is the same as the second derivative of x8 in this situation, however in your original post you had $$\bigg(\frac{dy}{dx}\bigg)^2$$ which is the square of the derivative, not a second derivative.

23. Jan 5, 2007

### Dr Game

wow.. I see how to do the second one.. its almost like a whole new order of operations to me... you derive first, then square it...

I'm very gracious for you spending your time and effort with me, thank you

24. Jan 6, 2007

### theperthvan

you screwed up cancelling the d's anyway even if it was a number.

25. Jan 31, 2007

### rbzima

Its in terms of derivatives. When you square y', you have y' squared. When you have (d^2 y)/(dx^2) you have the second derivative.

Suppose I have a position function of a train along a track, with the y axis telling me where it is with respect to time. y' would simply be the velocity at which it travels, and y" would simply be the acceleration of that train. Squaring velocity wouldn't give you acceleration. y" and y' are two completely different things.