Taking a time derivative of a function of 3 variables. I have a function of $3$ variables which are all functions of $t$.
$$x = \frac{v_1t-y}{\sqrt{(v_2/\dot{x})^2 -1}} \tag 1 $$ 
In the equation $v_1,v_2$ are constant and $x$ and $y$ are both function of $t$ (also $\dot{x}$ is $\frac{dx}{dt}$). I am trying to differentiate $(1)$ with respect to $t$, but I am not sure how to do this as there are three variables and an $\dot{x}$ already. I tried holding $x,y$ constant but that instinctively does not make sense as these change with respect to $t$.
 A: Hint:
If $y(t)$ and $\dot{x}(t)$ are two functions independent from $x(t)$
 you can use quotient rule and chain rule.
If $\dot{x}=\dfrac{dx}{dt}$, as I suppose, you have not a function but a differential equation that you can manipulate formally solving for $\dot{x}$ and find:
$$
\dot{x}=\dfrac{dx}{dt}=F(t,x(t),y(t))=\dfrac{v_2x}{\sqrt{(v_1t-y)+x^2}}
$$
But it seems not simple to solve.
A: I would first square both sides of the equation to get rid of the square root in the denominator: 
$$x^{2}=\frac {(v_{1}t-y)^{2}}{(\sqrt{(\frac {v_{2}}{\dot{x}})^{2}-1})^{2}}$$
$$x^{2}=\frac {v_{1}^{2}t^{2}-2v_{1}ty+y^{2}}{(\frac {v_{2}}{\dot{x}})^{2}-1}$$
Then rearrange it this way: 
$$((\frac {v_{2}}{\dot{x}})^{2}-1)x^{2}=v_{1}^{2}t^{2}-2v_{1}ty+y^{2}$$
$$\frac {v_{2}^{2}x^{2}}{\dot{x}^{2}}-x^{2}=v_{1}^{2}t^{2}-2v_{1}ty+y^{2}$$
$$\frac {v_{2}^{2}x^{2}-x^{2}\dot{x}^{2}}{\dot{x}^{2}}=v_{1}^{2}t^{2}-2v_{1}ty+y^{2}$$
$$v_{2}^{2}x^{2}-x^{2}\dot{x}^{2}=(v_{1}^{2}t^{2}-2v_{1}ty+y^{2})\dot{x}^{2}$$
$$v_{2}^{2}x^{2}=(v_{1}^{2}t^{2}-2v_{1}ty+y^{2})\dot{x}^{2}+x^{2}\dot{x}^{2}$$
$$v_{2}^{2}x^{2}=(v_{1}^{2}t^{2}-2v_{1}ty+y^{2}+x^{2})\dot{x}^{2}$$
$$\dot{x}^{2}=\frac {v_{2}^{2}x^{2}}{v_{1}^{2}t^{2}-2v_{1}ty+y^{2}+x^{2}}$$
$$\dot{x}=\sqrt \frac {v_{2}^{2}x^{2}}{v_{1}^{2}t^{2}-2v_{1}ty+y^{2}+x^{2}}$$
$$\dot{x}=\frac {v_{2}x}{\sqrt{v_{1}^{2}t^{2}-2v_{1}ty+y^{2}+x^{2}}}$$
$$\dot{x}=\frac {dx}{dt}=\frac {v_{2}x}{\sqrt{(v_{1}t-y)^{2}+x^{2}}}$$
This would be your answer since $\dot{x}=\frac {dx}{dt}$, hope I helped:)
