Why is the direction of propagation of $y = \sin(kx - \omega t) \quad \omega , k \gt 0$ toward $+X$ axis? I'm not sure if this question should be in Physics or here in Mathematics. Forgive me if it doesn't belong here.
My book says that the wave $y = \sin(kx -\omega t)$ travels in $+X$ direction. I tried deriving the speed of propagation, and I am getting the direction to be towards $-X$
We have, $$\frac{\partial y}{\partial t} = -\omega \cos(kx - \omega t)$$
Also, $$\frac{\partial y}{\partial x} = k \cos(kx - \omega t)$$
Dividing,
$$\frac{\partial x}{\partial t} = - \frac \omega k $$
This is speed of propagation of wave. Now I'm not sure if partial derivative of $x$ with $t$ gives speed of wave or not, but since the magnitude of speed is coming out to be $\frac \omega k$ which is the actual speed of a wave, I must be close.
The Problem
From my derivation, I get the speed of wave $= - \frac \omega k $, which means wave is travelling in $-X$ direction. My book says that the wave $y = \sin(kx - \omega t)$ travels in $+X$ direction with a speed of $\frac \omega k$. The wave $y = \sin(kx + \omega t)$ travels in $-X$ direction, which according to my calculation should go in $+X$
Why do I get the direction to be opposite? The magnitude of speed from my calculation is correct, but direction is coming out to be opposite.
 A: Let's first address another question: What does it mean for the wave to be propagating? In other words, how do I construct a useful definition for velocity?
One way to approach this, which is of interest to you, is by consdiering the phase velocity$^\dagger$,
$$v_p = \frac{\omega}{k}$$
We'll derive that in a second. First, let's discuss meaning. Phase velocity is how quickly a point of arbitrary phase propagates forward in time. Imagine a surfer riding an (ideal, gnarly) ocean wave to shore. Imagine that the surfer sticks to one point of the crest as he or she is carried forward. As the surfer does so, they pass one buoy and then another buoy whose relative distance is known. As an observer, you could measure that distance $\Delta x$ in some $\Delta t$ that the surfer moved through all while sticking to one spot on the wave, which is to say one constant $\phi_0$.
Now onto the derivation. For some constant phase, we have
$$\phi_0 = kx - \omega t$$
Differentiating w.r.t time, we get
$$ 0 = k\frac{dx}{dt} - \omega \implies \frac{dx}{dt} = \frac\omega k$$

$^\dagger$ There is more than one useful definition of velocity. Look up the difference between phase and group velocity for a more complete picture of wave motion.
A: Your mistake is in the calculation of the derivative of $x$ with respect to $t$ that is not he quotient of the partial derivatives, but that must be calculated using the derivation rule for an implicit function.
The velocity is the derivative $\frac{dx}{dt}$ of a function $\sin (kx-\omega t)=A$ that gives us the variation in time of the position $x$ of a point of the wafe with value $A$. This is done by:
$$
k \cos (kx-\omega t) \frac{dx}{dt}-\omega \cos (kx+\omega t)=0
$$
that gives:
$$
\frac{dx}{dt}=\frac{\omega}{k}
$$
A: To keep $kx-\omega t$ constant as $t$ increases, $x$ must increase as well. (Assuming that $k$ and $\omega$ are positive.) So your book is right, the wave travels in the positive $x$ direction.
More generally, for any function $f(u)$, the equation $y=f(kx-\omega t)$ describes a right-moving wave whose profile has the shape given by the function $f$. The level sets in the $(x,t)$ plane are the lines $kx-\omega t=\text{constant}$.
