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To solve the problem, we need to find \( \frac{dy}{dx} \) using implicit differentiation for the given equation:

\[
\sin\left(\frac{x}{y}\right) = \frac{1}{2}
\]

### Step 1: Differentiate both sides with respect to \(x\).

Using implicit differentiation, differentiate \( \sin\left(\frac{x}{y}\right) \) with respect to \(x\), while considering that \(y\) is a function of \(x\):

\[
\cos\left(\frac{x}{y}\right) \cdot \left(\frac{d}{dx} \left(\frac{x}{y}\right)\right) = 0
\]

Now, find \( \frac{d}{dx}\left(\frac{x}{y}\right) \) using the quotient rule:

\[
\frac{d}{dx}\left(\frac{x}{y}\right) = \frac{y \cdot 1 - x \cdot \frac{dy}{dx}}{y^2} = \frac{y - x \frac{dy}{dx}}{y^2}
\]

So, substituting this back, we get:

\[
\cos\left(\frac{x}{y}\right) \cdot \frac{y - x \frac{dy}{dx}}{y^2} = 0
\]

### Step 2: Solve for \( \frac{dy}{dx} \).

Since \( \cos\left(\frac{x}{y}\right) \neq 0 \), we can equate the numerator to zero:

\[
y - x \frac{dy}{dx} = 0
\]

This simplifies to:

\[
x \frac{dy}{dx} = y
\]

Thus,

\[
\frac{dy}{dx} = \frac{y}{x}
\]

The solution is:

\[
\frac{dy}{dx} = \frac{y}{x}
\]

Would you like further details or have any other questions?

### Related Questions:
1. What is implicit differentiation, and how does it differ from explicit differentiation?
2. How can we apply the chain rule in implicit differentiation?
3. What is the quotient rule, and how is it used in differentiation?
4. Why do we need to differentiate both sides of the equation with respect to \(x\)?
5. How can you solve implicit differentiation problems where multiple terms contain \(y\)?

### Tip:
When dealing with implicit differentiation, always remember to treat \(y\) as a function of \(x\), applying the chain rule appropriately.

Solve the problem using implicit differentiation for the equation sin(x/y) = 1/2 and find dy/dx.

Learn how to solve for dy/dx using implicit differentiation with the equation sin(x/y) = 1/2. This step-by-step solution covers key calculus concepts such as the chain rule, quotient rule, and trigonometric derivatives. Ideal for college-level calculus students.

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