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## Topic: Leibnitz notation in calculus (Read 715 times)previous topic - next topic

• stcordova
• Global Moderator
Re: Liebnitz notation in calculus
##### Reply #15 – January 22, 2018, 09:50:29 AM
There is probably a formal way to prove this relation as all the delta's go  to zero.  The problem is that in elementary calculus books, this proof is omitted!  For all the proof-based calculus book out there, this was one case where the proof would have been helpful, if only in the appendix or somewhere, rather than just letting the students remain in a state of confusion:

$\frac {\Delta \left ( \frac{\Delta y}{\Delta x} \right )}{\Delta z}=\frac {\Delta \left ( \frac{\Delta y}{\Delta z} \right )}{\Delta x}=\frac{\Delta \left ( \Delta y \right )}{\Delta x\Delta z}$

• stcordova
• Global Moderator
Re: Liebnitz notation in calculus
##### Reply #16 – January 22, 2018, 10:32:51 AM
So trying to deal with:
$\frac {\Delta \left ( \frac{\Delta y}{\Delta x} \right )}{\Delta z}=\frac {\Delta \left ( \frac{\Delta y}{\Delta z} \right )}{\Delta x}=\frac{\Delta \left ( \Delta y \right )}{\Delta x\Delta z}$

The right hand side of the equation relates to this.  One can see the h^2 correspondes to dx^2 (I guess):

$\displaystyle f''(x) = \lim_{h\to0} \frac {f(x+h)-2f(x)+f(x-h)}{h^2}$

The left hand side (I think) relates to:

$\displaystyle \frac {\frac{f(x+h)-f(x)}{h}-\frac{f(x)-f(x-h)}{h}}{h}$

I think coverting the "h"  to "delta x", plus some clean up,  will get the desired result to justify the Liebnitz notation.

It would have been helpful if they showed the connection in calculus text rather than just throwing Liebnitz notation in with not a lot of justification.

• stcordova
• Global Moderator
Re: Liebnitz notation in calculus
##### Reply #17 – January 22, 2018, 10:42:33 AM
I realized I went about things the hard way.  Cleaning up a bit, and taking out the "z" and replacing with "x".

$\frac {\Delta \left ( \frac{\Delta y}{\Delta x} \right )}{\Delta x} \approx \frac{\Delta \left ( \Delta y \right )}{\Delta x\Delta x}$

Note:

$\displaystyle \lim_{h\to0} \frac {\frac{f(x+h)-f(x)}{h}-\frac{f(x)-f(x-h)}{h}}{h}= \lim_{h\to0} \frac {f(x+h)-2f(x)+f(x-h)}{h^2}$

this relation has to just be recast to delta-X's and Y's, I think.

• stcordova
• Global Moderator
Re: Liebnitz notation in calculus
##### Reply #18 – January 22, 2018, 10:57:07 AM
Let me recast:

$\displaystyle \lim_{\Delta x\to0} \frac {\frac{f(x+\Delta x)-f(x)}{\Delta x}-\frac{f(x)-f(x-\Delta x)}{\Delta x}}{\Delta x}= \lim_{\Delta x\to0} \frac {f(x+\Delta x)-2f(x)+f(x-\Delta x)}{(\Delta x)^2}$

If I define:

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

and

$\Delta (\Delta y) = {f(x+\Delta x)-2f(x)+f(x-\Delta x)}$

then

$\displaystyle \lim_{\Delta x\to0} \frac {\Delta \left ( \frac{\Delta y}{\Delta x} \right )}{\Delta x} = \lim_{\Delta x\to0} \frac{\Delta \left ( \Delta y \right )}{\Delta x\Delta x}$

which justifies the notation

$\displaystyle \frac {d \left ( \frac{dy}{dx} \right )}{dx} = \frac{d \left ( dy \right )}{dxdx}=\frac {d^2y}{(dx)^2}=\frac {d^2y}{dx^2}$

• johnnyb
• Global Moderator
Re: Liebnitz notation in calculus
##### Reply #19 – January 22, 2018, 04:03:32 PM
Yes, it is a terrible tragedy that this never gets explained in textbooks!  What is even more of a tragedy is that, I believe, the notation itself is incorrect.  This is detectable if one actually takes $\frac{d\left(\frac{dy}{dx}\right)}{dx}$ seriously as a differential.

To perform this operation, you would actually need to use the quotient rule.  Doing so leads to a more expanded form of the second derivative:

$\frac{d^2 y}{dx^2} - \frac{dy}{dx}\frac{d^2 x}{dx^2}$

Now, this reduces to $\frac{d^2 y}{dx^2}$ in the case of x being an independent variable.  But, when x is not an independent variable, then the full expansion is needed.

If you have the full expansion, you can easily convert the second derivative of y with respect to x into the second derivative of x with respect to y using algebraic manipulations only, which is not possible with the traditional notation.

• Alan Fox