Can someone please explain why , where y is a function of x, is necessarily zero? Is it because holding x in place forces y' to never change?

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- April 16th 2013, 06:16 PM #1

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- April 16th 2013, 06:23 PM #2

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## Re: d(dy/dt)/dy

Hey phys251.

Using the chain rule we get d/dy (y'(x)) = dy'(x)/dy * dy/dy = 0 * 1 = 0. I'm not sure if this answer suffices or not to answer your question, but the chain rule is always a good tool when you have these kinds of derivatives.

- April 16th 2013, 06:48 PM #3

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- April 16th 2013, 06:53 PM #4

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- April 16th 2013, 06:56 PM #5

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- April 16th 2013, 07:03 PM #6

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- April 17th 2013, 12:00 AM #7

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- April 17th 2013, 01:13 AM #8

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- April 17th 2013, 04:19 AM #9

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## Re: d(dy/dt)/dy

Of course, if y is not a function of x or related to x then y'(x)=0 and there is no need for further discussion !

The aim of my first post at arousing the controversy because the wording of the question was ambiguous.

Introducing the partial derivative of (dy/dx) relatively to y suppose that dy/dx might be function of y, i.e.: dy/dx = f(x,y). There are many different manner to write f(x,y) and all are equal.

For example, in the case y(x)=x²

f(x,y) = 2x, or f(x,y)=2sqrt(y), or f(x,y)=x+sqrt(y), or f(x,y)=2y/x, or etc.

The partial derivative relatively of y is different for each one of these functions f(x,y). So, without more specification in the wording, the question is ambiguous.

Among many possibilities, if we say : y(x) relates y to x as well as the inverse function x(y), then dy/dx is function of x which is function of y. Hence dy/dx is function of y. On this point of view, the formula for d(dy/dx)/dy is shown in attachment.