# Partial derivative of $v$ w.r.t. $x$ in Lagrangian dynamics [duplicate]

In Lagrangian dynamics, when using the Lagrangian thus:

$$\frac{d}{dt}(\frac{\partial \mathcal{L} }{\partial \dot{q_j}})- \frac{\partial \mathcal{L} }{\partial q_j} = 0$$

often we get terms such as the one below which result in zero:

example 1: $$\frac{\partial \dot{q}}{\partial q} = 0$$ example 2: $$\frac{\partial q}{\partial \dot{q}} = 0$$

I wasn't sure why but this is the closest I have come to reasoning it out:

$$\frac{\partial \dot{q}}{\partial q} = \frac{\partial \dot{q}}{\partial t}\frac{\partial t}{\partial q} = 0$$

because

$$\frac{\partial t}{\partial q} = 0$$

and similar for example 2. Is this correct?

## marked as duplicate by AccidentalFourierTransform, Qmechanic♦ lagrangian-formalism StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); May 11 '17 at 15:50

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