Calculus of Variation

Because there is no algebraic relation between a function and its derivative.

This is why you need boundary conditions to solve differential equations.

 

Sorry, but this is a bogus answer. A function may depend on another function non-algebraically, and that is perfectly fine as far as functional dependency goes. Not to mention that the dependency may perfectly well be algebraic.

The real reason is that we use the partial derivatives to obtain an expression for the difference ## F(z + \Delta z, y + \Delta y, x) - F(z, y, x) ##, which is approximately ## F_z \Delta z + F_y \Delta y ## when ##\Delta z## and ##\Delta y## are sufficiently small. This expression is true generally, and is true when ## z ## represents the derivative of ## y ## - all it takes is that the variations of both must be small enough. If ## y = f(x) ##, its variation is ## \delta y = \epsilon g(x) ##, and ## \delta y' = \epsilon g'(x)##. If ## \epsilon ## is small enough, then using the result above, ## F((y + \delta y)', (y + \delta y), x) - F(y', y, x)) \approx \epsilon F_{y'}g'(x) + \epsilon F_y g(x) ##, where ##F_{y'}## is just a fancy symbol equivalent to ##F_z##, meaning partial differentiation with respect to the first argument. Then we use integration by parts and convert that to ## \epsilon (-(F_{y'})' + F_y) g(x)##. Observe that we do use the relationship between ## y ## and ## y' ## in the final step.

 

Would the following also be correct reasoning?

We want to find the least action for:

##S = \int_{x_1}^{x_2} f(y,y',x) \, dx##

While this may look as though y, y' and x are simple independent variables, since we are actually looking for the function f(x) that provides this least action, what this notation really means is this:

##S = \int_{x_1}^{x_2} f[y(x), \frac d {dx} y(x), x] \, dx##

So y and y' are not truly independent.