Can $A \sin (kx - \omega t)$ be used for longitudinal wave? 

Question: One particle in the medium has its equilibrium position at $x = 1.00 \ m $. Show that the speed of this particle at $t=0.882 \ s$ is $4.9 \ m/s$.

I attempted at using the formula
$$y=A \sin(kx- \omega t)$$
and then I tried to use
$$\frac{dy}{dt}=\frac{dy}{dx} \times \frac{dx}{dt}$$
to evaluate how $y$ changes with respect to time (which is what the question asks).
But when I plug in $x=1.00$ and $t=0.882$, then I get a negative $\frac{dy}{dx}$, but that contradicts with the fact that in the graph the tangent has a positive slope.
So, the question: Is it wrong to use $y = A \sin (kx - \omega t)$ in this question where the wave is longitudinal?
 A: You can use $y(x,t)=A\sin(kx-\omega t+\phi)$ for longitudinal waves. But understand its meaning: $y$ denotes the longitudinal displacement of the particle whose equilibrium position is $x$, at time $t$.
The problem wants you to evaluate the speed of the particle whose equilibrium position is $x=1$ (let's label the particle as "$1$" since there is only one particle whose equilibrium position is $x=1$). Therefore, $f_1(t)\equiv y(1,t)=A\sin(k-\omega t+\phi)$ gives the displacement of "$1$" and $v_1(t)\equiv \frac{df}{dt}(t)=-A\omega \cos(k-\omega t + \phi)$ gives the velocity of "$1$" at time $t$.
The values of $A$, $k$, $\omega$, and $\phi$ can be determined using the given graphs.

Response to comments are made here

You directly differentiated with respect to time (maybe partial derivative?) ... 

$\frac{\partial y}{\partial t}(x,t)$ gives the velocity of the particle whose equilibrium position is $x$, at time $t$. I defined a function called $f_x(t)=y(x,t)$ to specifically denote the displacement of the particle whose equilibrium position is $x$. But the essence is the same: $\frac{\partial y}{\partial t}(x,t)=\frac{df_{x}}{dt}(t)$.

What is the reason why the use of chain rule here is not valid? 

Chain rule makes sense in cases such as the one shown below. Here, $y$ is a function of $h$ which in turn is a function of $x$ and one would like to know how much $y$ changes when one varies $x$.
$$y=y(h(x)) \Rightarrow \frac{dy}{dx}=\frac{dy}{dh} \times \frac{dh}{dx}$$
To the contrary, in our wave equation, $x$ and $t$ are independent variables. The $x$ in $y(x,t)$ identifies the particle whose displacement we would like to determine. In other words, $x$ identifies the particle of interest. If you specify $x=1$, the function $y$ thinks "Oh, he's referring to the particle whose equilibrium position is $x=1$. Cool. At what time would you like to know its displacement?". Time is an additional piece of information that is not determined by $x$.
