Full time-derivative, <PERSON> brackets and <PERSON>'s equations (classical mechanics)
While studying <PERSON> brackets in classical mechanics and the derivation of $\dot{q_j}={q_j,H}$ and $\dot{p_j}={p_j,H}$ form of <PERSON>'s equations I encountered a surpsing identity, which led me to think that maybe I got something wrong about full time-derivative, which is as follows:
$$ \frac{d\,f(q_1,q_2,\dots,q_N,p_1,p_2,\dots,p_N,t)}{dt}=\sum_{j=1}^{N}\left( \frac{\partial f}{\partial q_j} \underbrace{\frac{\partial q_j}{\partial t}}{=\dot{q_j}=\frac{\partial H}{\partial p_j}}+ \frac{\partial f}{\partial p_j} \underbrace{\frac{\partial p_j}{\partial t}}{=\dot{p_j}=-\frac{\partial H}{\partial q_j}} \right)+\frac{\partial f}{\partial t} ={f,H}+\frac{\partial f}{\partial t} $$
Now if for example I use function $f$ as $q_k$, by saying: $f(q_1,q_2,\dots,q_N,p_1,p_2,\dots,p_N,t)=q_k$ then I get following:
$$ \frac{dq_k}{dt}=\sum_{j=1}^{N}\left( \underbrace{\frac{\partial q_k}{\partial q_j}}{=\delta{kj}} \frac{\partial H}{\partial p_j}- \underbrace{\frac{\partial q_k}{\partial p_j}}{=0} \frac{\partial H}{\partial q_j} \right)+\frac{\partial q_k}{\partial t} ={q_k,H}+\frac{\partial q_k}{\partial t} = \underbrace{\frac{\partial H}{\partial p_k}}{=\dot{q_k}} + \underbrace{\frac{\partial q_k}{\partial t}}_{\stackrel{?}{=}\dot{q_k}}\stackrel{?}{=}2\dot{q_k} $$
If I cross out the $\frac{\partial q_k}{\partial t}$ on both sides of equation ${q_k,H}+\frac{\partial q_k}{\partial t}=\frac{\partial H}{\partial p_k}+\frac{\partial q_k}{\partial t}$ then I recover the <PERSON>'s equation $\dot{q_k}={q_k,H}$. But if I don't do this, and go forward with the $2\dot{q_k}$ that appears at the end I get this very surprising identity:
$$ \frac{dq_k}{dt}\stackrel{?}{=}2\dot{q_k} $$
or written another way:
$$ \frac{dq_k}{dt}\stackrel{?}{=}2\frac{\partial q_k}{\partial t} $$
My question is following: is this really true? If not, then what have I done wrong? If yes, then why it is not mentioned anywhere in the textbooks - wouldn't that be some other way to find the $\dot{q_k}$?
Note:The derivation goes exactly the same for $\dot{p_k}$.
EDIT: Thanks to your answers I marked now the wrong equalities with $\stackrel{?}{=}$ since I hate to see incorrect math typed out. But I still wanted to preserve this question as how it was written at first. 955