Implicit differentiation issue I am trying to implicitly differentiate $$\sin(x/y) = 1/2 $$
The solution manual says
Step 1. $$\cos(x/y)\cdot\frac{y-x\frac{dy}{dx}}{y^2} = 0 $$
But I don't understand how they arrive at this next part:
Step 2. $$y-x\frac{dy}{dx}=0$$
Is $\cos(x/y) = y^2$?
 A: One approach, already mentioned in a posted answer here, says that the cosine cannot be $0$ when the sine is $1/2$, so if $\cos(\text{something})\cdot (\text{something}) = 0$, then the second "something" must be $0$.
Another approach seems simpler, since it avoids differentiating any trigonometric functions and applying the chain rule to that differentiation.  Just observe that
$$
\frac xy = \arcsin\frac12.
$$
Differentiate the left side using the quotient rule and the chain rule.  When you differentiate the right side, you get $0$ since the right side is constant, and all references to trigonometric functions disappear.
A: The idea is that when $ab = 0$, you have that $a = 0$ or $b=0$ as solutions to the equation (or possibly both). Here, we have $\dfrac{1}{y^2}\cos\left(\dfrac{x}{y}\right) = 0$ or $y-xy' = 0$.
Since $y$ cannot be "infinity", from the first solution we have that $\cos\left(\dfrac{x}{y}\right) = 0$ which says $\frac{x}{y} = \frac{\pi}{2},\frac{3\pi}{2},\cdots$. So really, $y$ is a linear function of $x$. But you can check that none of the above values I gave satisfy $\sin\left(\dfrac{x}{y}\right) = \dfrac{1}{2}$ (in fact it is equal to $\pm 1$ for these values). Check this yourself.
So that leaves us with $y-xy' = 0$. Is it clear now?
A: applying chain rule at $LHS$ and differentiating. let $\frac{x}{y}=t$$$\frac{d}{dt}\cos (t)\frac{dt}{dx}=0$$this yields the answer
