What is the difference between d^2y/dx^2 and dy^2/dx^2?

[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...

 

[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.

 

[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.

 

[Dr Game]

I can understand that.

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

alright.. let's say was y=x^4 how are the two different

 

[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.

 

[d_leet]

I can understand that.

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

alright.. let's say was y=x^4 how are the two different

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.

 

[Dr Game]

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

 

[d_leet]

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

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.

 

[Hurkyl]

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.

 

[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.

 

[d_leet]

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 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.

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.

 

[gabee]

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.

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

 

[Dr Game]

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

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

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

 

[d_leet]

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

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

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

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.

 

[Hurkyl]

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.

 

[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

 

[d_leet]

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

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.

 

[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 textbook, 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...

 

[gabee]

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

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

I just read the section in my textbook, and this opened up everything, I see things clearly now...

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.

 

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

 

[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 $${d^2 \over dx^2} y^2$$.

 

[d_leet]

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 textbook, and this opened up everything, I see things clearly now...

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.

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

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

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

you derive it because it is an operator...

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

This part is correct.

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...

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.

 

[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

 

[theperthvan]

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

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

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

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

 

[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.

 

[Mr.4]

Well, I was really lucky to have some of the best Math teachers and books to grow up with. This is what I've understood from them

dy/dx or the first derivative of a function means the slope of the graph of the function at a point. For eg:

if y= x^3
then dy/dx = 3x^2

So the slope at any point, say (2,8) (x=2, y= 2^3=8) is given by 3*2^2=12.

This hold for all the points on the curve. Thus if we plot the graph for dy/dx against x we get another continuous function (This needn't always be true i.e the original function can be continuous but its derivative non-discontinuous). So we can treat dy/dx as another function.

If we want to find the slope at any point of this new function we simply have to differentiate it again. i.e d/dx(dy/dx) (Uhhh... u get the expression!)

Theoretically the 1st derivative means rate of change of y with respect to x and 2nd derivative means rate of change of rate of change of y w.r.t x and 3rd derivative means rate of change of rate of change of rate of change of y w.r.t z and so on.

You may look read the chapter on Maxima and Minima with this in mind. There we usually check whether the second derivative is +ve or -ve corr. to a minima and maxima respectively. But that's another story

 

[HallsofIvy]

There is no such thing as "d²" by itself like that. I suspect that is the basic problem. We can define the differentials, dx and dy, but not "d²y" or "dx²".

 

[guitarra]

this is the second derivative of y with respect to x

 

[ChaoticLlama]

I can't believe no one stated this before.

lim ∆x -> 0 ∆y/∆x = dy/dx

You cannot solve for ∆, any more than you can solve for d.

Your only issue lies in that you don't understand BEDMAS, or whatever mnemonic you use to remember order of operations.