Proof of Hamiltonian equations

[Daaavde]

So, I should prove that:
$- \frac{\partial H}{\partial q_i} = \dot{p_i}$
And it is shown that:
$- \frac{\partial H}{\partial q_i} = - p_j \frac{\partial \dot{q_j}}{\partial q_i} + \frac{\partial \dot{q_j}}{\partial q_i} \frac{\partial L}{\partial \dot{q_j}} + \frac{\partial L}{\partial q_i} = \dot{p_i}$
Where the first two terms delete each other $(\frac{\partial L}{\partial \dot{q_j}} = p_j)$ and the third one is equal to $\dot{p_i}$ because of the Lagrange equation.

The problem is that when I take the partial derivative of $H = \sum \dot{q_i}p_i - L$, I get:
$- \frac{\partial H}{\partial q_i} = - p_j \frac{\partial \dot{q_j}}{\partial q_i} - \dot{q_j} \frac{\partial p_j}{\partial q_i} + \frac{\partial L}{\partial q_i} = \dot{p_i}$
Because I derive a product.

Now, my second term is completely different (even the sign doesn't match). Why is that?

 

[voko]

Have a look here: https://www.physicsforums.com/threads/hamiltonian-mechanics-h-t.769278/#post-4843651

 

[Daaavde]

Sorry, I can't see how that answers the question.
There are no time derivatives here or maybe I overlooked something.

 

[voko]

Ignore the time part. Make sure you understand the technique that yields the differential of the Hamiltonian that only has the differentials of the canonical variables (and time). From that, the partial derivatives w.r.t. canonical variables follow easily.