TITLE: how is this differentiation possible?
QUESTION [5 upvotes]: A particle moves in the plane xy with constant acceleration a directed along the negative y–axis. The equation of motion of the particle has the form (where p and q are constants)
$$y = px –{qx}^2$$
in my book they differentiated both sides with respect to time and solved it as
$$\frac {dy}{dt}=p\frac{dx}{dt}+q\cdot 2x \cdot\frac{dx}{dt}$$
they later took $\frac{dx}{dt}$ as velocity in x-axis and solved it....
my question is that we dont know the relation of x with t.so how could we do $\frac{d}{dt}{qx}^2$ and get $q\cdot 2x\cdot\frac{dx}{dt}$ after differentiating?
i have tried asking the question before and my ques got closed as i didnt know mathjax.so plz dont downvote and tell me what i can do to better edit the question.

REPLY [3 votes]: By the chain rule,
$$
\frac{d\left(qx^2\right)}{dt}=\frac{d\left(qx^2\right)}{dx} \cdot \frac{dx}{dt} \, .
$$
So
$$
\frac{d\left(qx^2\right)}{dt}=2qx\cdot\frac{dx}{dt}\, . \tag{*}\label{*}
$$
It is true that we don't know how $x$ is related to $t$, which is why it is not possible to simplify the expression $\frac{dx}{dt}$ any further. But regardless of how $x$ is related to $t$, the equation $\eqref{*}$ holds.