1. ## d(dy/dt)/dy

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

2. ## 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.

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

dy/dy is obvious, but I still don't get why dy'(x)/dy = d(dy/dx)/dy is zero.

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

Its a function of x so its like differentiating a constant.

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

Originally Posted by chiro
Its a function of x so its like differentiating a constant.
So, because the original equation was a partial, x remains constant, ergo y'(x) is constant too? If so, the partial with respect to y is definitely zero.

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

That is correct: since the expression doesn't involve y its like differentiating with respect to a constant term (which is zero).

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

Hi !

Sorry to say : I don't agree with your line of reasoning
Just look at the example in attachment and see that the result is not equal to zero.
I let you think about for a while before giving the correct answer.

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

To finish what JJacquelin started: if y is not a function of x or related to x, then my post holds but yes y is related to x (so obvious once it is pointed out).

You can use the inverse function theorem to make more specific claims.

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

Originally Posted by chiro
To finish what JJacquelin started: if y is not a function of x or related to x, then my post holds but yes y is related to x (so obvious once it is pointed out).
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.

10. ## 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
$$\frac{\frac{dy}{dx}\cdot \frac{\partial}{\partial y}\frac{dy}{dx}}{\sqrt{1+{\frac{dy}{dx}}^2}}$$