# Why y, y' (derivative of y), x are independent?

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1. May 25, 2015

### SSDdefragger

In calculus of variations when we solve Euler's equation we always do think of y, x and y' as independent variables.
In thermodynamics we think that different potentials have totally different variables
I don't understand why the slope of the function is not directly dependent on function itself.

2. May 25, 2015

### Orodruin

Staff Emeritus
It is directly dependent on the function. This is why you get one differential equation as the Euler equation and not several.

However, when computing the variation of a function which depends on several variables, you need to take the partial derivative wrt each argument separately and multiply by the inner derivative. That the arguments are not independent manifests itself by the inner derivatives not being independent.

3. May 25, 2015

### SSDdefragger

Please can you explain this in more plain English, I don't quite follow.

4. May 25, 2015

### Orodruin

Staff Emeritus
You have a function of the form $f(a,b,c)$ where you are going to plug in $a = y(x)$, $b = y'(x)$, and $c = x$. If you want to know how this changes when you let $y(x) = y_0(x) + \epsilon y_1(x)$, you do the following:
$$\frac{df}{d\epsilon} = \frac{\partial f}{\partial a} \frac{da}{d\epsilon} + \frac{\partial f}{\partial b} \frac{db}{d\epsilon} + \frac{\partial f}{\partial c} \frac{dc}{d\epsilon}.$$
Here, $a$, $b$, and $c$ are just placeholders for the functions that are going to go into this. What $\partial f/\partial a$ really means is "the derivative of $f$ with respect to the first argument. In order not to have to write down dummy variables like this, we usually just write $f(y,y',x)$ and by $\partial f/\partial y$ we mean the derivative of $f$ with respect to its first argument and so on. Inserting $a = y(x)$, $b = y'(x)$, and $c = x$ gives you the correct result and $da/d\epsilon = y_1$, $db/d\epsilon = y'_1$, and $dc/d\epsilon = 0$. The fact that $y'_1$ is not independent of $y_1$ encodes the fact that $y'$ is not independent of $y$.

To take an analogy to usual multivariable calculus, consider the function $f(x,y) = xy$. If we let $x$ and $y$ depend on a parameter $t$, we find that
$$\frac{df}{dt} = x' \partial_x f + y' \partial_y f = x' y + y' x.$$
If we now assume that $y = x^2$, we would find $y' = 2xx'$ and inserting this into the above would give
$$\frac{df}{dt} = x' x^2 + 2xx' x = 3 x^2 x',$$
the very same result as we would get if we had inserted $y = x^2$ into $f$ before differentiating, so it is consistent with $y$ being dependent on $x$.