Can someone please explain why $\displaystyle \frac{\partial}{\partial y}(y'(x))$, where y is a function of x, is necessarily zero? Is it because holding x in place forces y' to never change?

Results 1 to 10 of 10

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

- Joined
- Jun 2012
- From
- Georgia
- Posts
- 220
- Thanks
- 55

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

- Joined
- Sep 2012
- From
- Australia
- Posts
- 6,596
- Thanks
- 1712

## 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.

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

- Joined
- Jun 2012
- From
- Georgia
- Posts
- 220
- Thanks
- 55

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

- Joined
- Sep 2012
- From
- Australia
- Posts
- 6,596
- Thanks
- 1712

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

- Joined
- Jun 2012
- From
- Georgia
- Posts
- 220
- Thanks
- 55

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

- Joined
- Sep 2012
- From
- Australia
- Posts
- 6,596
- Thanks
- 1712

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

- Joined
- Aug 2011
- Posts
- 252
- Thanks
- 61

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

- Joined
- Sep 2012
- From
- Australia
- Posts
- 6,596
- Thanks
- 1712

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

- Joined
- Aug 2011
- Posts
- 252
- Thanks
- 61

## 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)=x

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.

- Nov 28th 2016, 08:38 PM #10

- Joined
- Nov 2016
- From
- United States
- Posts
- 1

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

I have this question too. I am looking at the Euler-Lagrange equation:

$$\frac{{\partial F}}{{\partial y}} - \frac{d}{dx}\frac{{\partial F}}{{\partial y'}}=0$$

where $F=F(y',y,x)$

As an example, I'm looking at a simple case of minimising arc length:

$$\int_{x_1}^{x_2}\sqrt{1+{\frac{dy}{dx}}^2}dx$$ where $y=y(x)$, and $F$ is everything inside the integral.

I'm trying to understand why the $\frac{\partial F}{\partial y}$ term is supposed to be 0 when y is a function of x. When computing $\frac{\partial F}{\partial y}$ I get

$$\frac{\frac{dy}{dx}\cdot \frac{\partial}{\partial y}\frac{dy}{dx}}{\sqrt{1+{\frac{dy}{dx}}^2}}$$

Click on a term to search for related topics.