1
$\begingroup$

Suppose I have a function $f = xy$. A partial derivative of $f$ with respect to $x$ implies holding $y$ constant:

$$ \frac{\partial f}{\partial x} = y $$

Does this mean that in order to evaluate this derivative, $y$ cannot depend on $x$? For example, if $y = x$ then

$$ \frac{\partial f}{\partial x} = \frac{\partial (x^2)}{\partial x} = 2x $$

Which is inconsistent with the first calculation of $\frac{\partial f}{\partial x}$.

If $y$ indeed cannot depend on $x$, then how does the Lagrangian formalism of classical mechanics make sense? The Lagrangian is a function of $q$ and $\dot{q}$, and when we evaluate $\frac{\partial L}{\partial \dot{q}}$ we 'ignore' all of the $q$ dependence, even though $\dot{q}$ is a function of $q$.

$\endgroup$
  • 2
    $\begingroup$ possible duplicate of Why does calculus of variations work? $\endgroup$ – Danu Jul 4 '14 at 19:18
  • $\begingroup$ Normally, the second way of calculating would be correct. However, in the Lagrangian formalism the velocity and position still appear as independent variables; they are not quite varied independently, as explained in the question I linked in the above comment. $\endgroup$ – Danu Jul 4 '14 at 19:49