Differentiate the following functions w.r.t. x.
`sin^2 (3x -2)`

Answer

To differentiate the function \( y = \sin^2(3x - 2) \) with respect to \( x \), we will use the chain rule. Here’s a step-by-step solution:

### Step 1: Identify the outer and inner functions
The function can be expressed as:
\[ y = (\sin(3x - 2))^2 \]
Here, the outer function is \( u^2 \) where \( u = \sin(3x - 2) \), and the inner function is \( u = \sin(3x - 2) \).

### Step 2: Differentiate the outer function
Using the chain rule, the derivative of \( y = u^2 \) with respect to \( u \) is:
\[ \frac{dy}{du} = 2u \]
Substituting back for \( u \):
\[ \frac{dy}{du} = 2\sin(3x - 2) \]

### Step 3: Differentiate the inner function
Now we need to find the derivative of \( u = \sin(3x - 2) \) with respect to \( x \):
\[ \frac{du}{dx} = \cos(3x - 2) \cdot \frac{d}{dx}(3x - 2) \]
The derivative of \( 3x - 2 \) is \( 3 \), so:
\[ \frac{du}{dx} = \cos(3x - 2) \cdot 3 = 3\cos(3x - 2) \]

### Step 4: Apply the chain rule
Now, we can apply the chain rule:
\[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \]
Substituting the derivatives we found:
\[ \frac{dy}{dx} = 2\sin(3x - 2) \cdot 3\cos(3x - 2) \]
This simplifies to:
\[ \frac{dy}{dx} = 6\sin(3x - 2)\cos(3x - 2) \]

### Step 5: Use the double angle identity
Using the trigonometric identity \( 2\sin A \cos A = \sin(2A) \), we can rewrite the expression:
\[ \frac{dy}{dx} = 3\sin(2(3x - 2)) \]
This simplifies to:
\[ \frac{dy}{dx} = 3\sin(6x - 4) \]

### Final Answer
Thus, the derivative of \( y = \sin^2(3x - 2) \) with respect to \( x \) is:
\[ \frac{dy}{dx} = 3\sin(6x - 4) \]

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