
#19
Jan405, 09:29 AM

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P: 11,866

[tex] \frac{d}{dx}\frac{dy}{dx}=\frac{d^{2}y}{dx^{2}} [/tex] Your notation makes no sense. The derivative of "n"th order wrt to "x" of the function "y" in the notation of Joseph Louis Lagrange is [tex] y^{(n)}(x) [/tex] In the notation of Gottfried Wilhelm Leibniz the same "animal" is [tex] \frac{d^{n}y(x)}{dx^{n}} [/tex] and it can be seen as appying the operator [itex] \frac{d}{dx} [/itex] "n" times on the function "y(x)". Daniel. 



#20
Jan405, 09:38 AM

P: 181

Hmm. I've a hard time understanding this. Can you point me in right direction? 



#21
Jan405, 09:53 AM

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P: 11,866

Okay,let's assume for simplicity that our functions depend only on variable and they are infinite times derivable on an arbitrary interval.
Then,we adopt Lagrange's notation and we say that the first order derivative of the function called y(x) is [tex] y'(x) [/tex].We can look on it as applying a linear operator called derivative which we'll denote by "'"(prime).So applying this operator on the function y(x),we get the derivative of the function "y(x)":[itex] y'(x) [/tex].The same operator (called 'derivative') is denoted by G.W.Leibniz as [itex] \frac{d}{dx} [/itex],since the result is the same (namely the first derivative of 'y' wrt to 'x'),no matter the notation,we can say [tex] y'(x)=\frac{d}{dx}y(x)=\frac{dy(x)}{dx} [/tex] For the second derivative,u apply this operator on the first derivative and u get,in the Lagrange notation: [tex] [y'(x)]'=y''(x)=y^{(2)}(x) [/tex] ,and in the Leibniz notation [tex] \frac{d}{dx}[\frac{dy(x)}{dx}]=\frac{d^{2}y(x)}{dx^{2}} [/tex] You equate the results,since they represent the same thing;the secondorder derivative of the function "y(x)" wrt to "x". [tex] y^{(2)}(x)=\frac{d^{2}y(x)}{dx^{2}} [/tex] And the same u do for every order of the derivative.For"n" [tex] y^{(n)}(x)=\frac{d^{n}y(x)}{dx^{n}} [/tex] Daniel. 



#22
Feb612, 05:42 AM

P: 26

"dx" is inifinitesimal but I'm confused that it's dx>0 or dx<0 ??




#23
Feb612, 05:48 AM

P: 4,570

You could make dx negative, but by doing that you have to adjust the other definitions even if you want to consider how the tangent changes when you are going 'backwards' (dx < 0). In numerical calculus we do have situations where we consider these kinds of things, but for normal calculus its almost always assumed that dx is an infinitesimal quantity with dx > 0 



#24
Feb612, 06:36 AM

P: 26

Okay, thanks.
But in this notation [itex]\int^{b}_{a}{f(x).dx}[/itex] I know it means area under the curve on interval [a,b] what if I take dx<0 how can I explain it? I mean I can take limit (lim [itex]\frac{Δy}{Δx}[/itex] as x approximates either side of xaxis (Δx→0[itex]^{+}[/itex] and Δx→0[itex]^{}[/itex]) I realize that if dx>0 or dx<0 at derivative of a function but I'm not sure about integral?? 



#25
Feb612, 07:37 AM

Sci Advisor
P: 1,716

A differentiable function can be approximated by the first term in its Taylor series
y = y0 + dxy' This approximation gets increasingly accurate as dx approaches zero. So the ratio (yy0)/dx is arbitrarily close to y' I think of dy as the change in y for a small change in x,dx. In physics books you will find this way of looking at it as well. 



#26
Feb612, 10:53 AM

P: 26

In Physics, we know that what "Δx" means. (Δx=x[itex]_{2}[/itex]x[itex]_{1}[/itex]) It can also be Δx>0 or Δx<0.In Calculus, Leibniz Notation shows dx=Δx and dy≠Δy I got that part why it's so.But if "Δx" can be both Δx>0 and Δx<0, for dx and dy it must be the same (dx,dy>0 or dx,dy<0) right??
Finally, in my opinion.We use from right and from left derivative to recognize that the function is differentiable.If we have a differentiable function, don't need to do this step.Therefore, Leibniz thought that "dx" infinitesimal number (dx=[itex]\frac{1}{∞}[/itex]>0) and used it a differentiable funtion, so that he didn't need to approximate the funtion to observe whether it's differentiable or not and he admitted his notation to find derivative of a funtion which doesn't need to be observed whether it's differentiable or not. Is that all right? 



#27
Feb612, 12:14 PM

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P: 1,716





#28
Feb612, 02:33 PM

P: 26

dx=Δx??




#29
Feb612, 08:02 PM

P: 4,570




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