Differentiate the following using Implicit Differentiation:  sin 𝑥𝑥 cos(𝑦𝑦 − 1) = 1 2

We are given the equation:

\[
\sin(x) \cos(y - 1) = \frac{1}{2}
\]

We will use implicit differentiation to differentiate both sides of the equation with respect to \( x \), treating \( y \) as a function of \( x \), i.e., \( y = y(x) \).

### Step 1: Differentiate the left-hand side

Using the product rule for differentiation (since we have a product of two functions), we first differentiate \( \sin(x) \) and \( \cos(y - 1) \) separately.

\[
\frac{d}{dx}[\sin(x) \cos(y - 1)] = \frac{d}{dx}[\sin(x)] \cdot \cos(y - 1) + \sin(x) \cdot \frac{d}{dx}[\cos(y - 1)]
\]

#### Differentiating \( \sin(x) \):

\[
\frac{d}{dx}[\sin(x)] = \cos(x)
\]

#### Differentiating \( \cos(y - 1) \):

We need to apply the chain rule here. Since \( y = y(x) \), we get:

\[
\frac{d}{dx}[\cos(y - 1)] = -\sin(y - 1) \cdot \frac{dy}{dx}
\]

Now we combine the differentiated parts:

\[
\frac{d}{dx}[\sin(x) \cos(y - 1)] = \cos(x) \cdot \cos(y - 1) + \sin(x) \cdot [-\sin(y - 1)] \cdot \frac{dy}{dx}
\]

### Step 2: Differentiate the right-hand side

The derivative of the constant \( \frac{1}{2} \) is simply 0:

\[
\frac{d}{dx}\left(\frac{1}{2}\right) = 0
\]

### Step 3: Set up the differentiated equation

So, we have:

\[
\cos(x) \cos(y - 1) - \sin(x) \sin(y - 1) \frac{dy}{dx} = 0
\]

### Step 4: Solve for \( \frac{dy}{dx} \)

Rearrange the equation to isolate \( \frac{dy}{dx} \):

\[
-\sin(x) \sin(y - 1) \frac{dy}{dx} = -\cos(x) \cos(y - 1)
\]

Dividing both sides by \( -\sin(x) \sin(y - 1) \) gives:

\[
\frac{dy}{dx} = \frac{\cos(x) \cos(y - 1)}{\sin(x) \sin(y - 1)}
\]

This is the derivative \( \frac{dy}{dx} \) using implicit differentiation.

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Do you want details or have any questions? Here are 5 questions related to this:

1. How does implicit differentiation handle multiple variables?
2. What is the product rule and how is it applied here?
3. Can you always solve for \( \frac{dy}{dx} \) explicitly after implicit differentiation?
4. Why is the chain rule necessary when differentiating \( \cos(y - 1) \)?
5. What are the common challenges in implicit differentiation?

**Tip:** Always use the chain rule when differentiating terms involving \( y \) since it's treated as a function of \( x \).

Differentiate the following using Implicit Differentiation: sin(x) cos(y - 1) = 1/2

Learn how to use implicit differentiation to solve the equation sin(x) cos(y - 1) = 1/2. This problem involves the product rule, chain rule, and trigonometric functions. Suitable for students in Grades 11-12 or early college-level calculus.

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