`y=sinxcdotsin2xcdotsin4xcdotsin8x`

Answer

To find the derivative of the function \( y = \sin x \cdot \sin 2x \cdot \sin 4x \cdot \sin 8x \), we will use logarithmic differentiation. Here’s a step-by-step solution:

### Step 1: Take the logarithm of both sides
We start by taking the natural logarithm of both sides:
\[ \log y = \log(\sin x \cdot \sin 2x \cdot \sin 4x \cdot \sin 8x) \]

### Step 2: Use the property of logarithms
Using the property of logarithms that states \( \log(m \cdot n) = \log m + \log n \), we can expand the right-hand side:
\[ \log y = \log(\sin x) + \log(\sin 2x) + \log(\sin 4x) + \log(\sin 8x) \]

### Step 3: Differentiate both sides
Now we differentiate both sides with respect to \( x \):
\[ \frac{d}{dx}(\log y) = \frac{d}{dx}(\log(\sin x)) + \frac{d}{dx}(\log(\sin 2x)) + \frac{d}{dx}(\log(\sin 4x)) + \frac{d}{dx}(\log(\sin 8x)) \]

Using the chain rule on the left side:
\[ \frac{1}{y} \frac{dy}{dx} = \frac{1}{\sin x} \cdot \cos x + \frac{1}{\sin 2x} \cdot 2 \cos 2x + \frac{1}{\sin 4x} \cdot 4 \cos 4x + \frac{1}{\sin 8x} \cdot 8 \cos 8x \]

### Step 4: Simplify the right side
This gives us:
\[ \frac{dy}{dx} = y \left( \frac{\cos x}{\sin x} + \frac{2 \cos 2x}{\sin 2x} + \frac{4 \cos 4x}{\sin 4x} + \frac{8 \cos 8x}{\sin 8x} \right) \]

### Step 5: Substitute back for \( y \)
Since \( y = \sin x \cdot \sin 2x \cdot \sin 4x \cdot \sin 8x \), we can substitute back:
\[ \frac{dy}{dx} = \sin x \cdot \sin 2x \cdot \sin 4x \cdot \sin 8x \left( \cot x + 2 \cot 2x + 4 \cot 4x + 8 \cot 8x \right) \]

### Final Answer
Thus, the derivative \( \frac{dy}{dx} \) is:
\[ \frac{dy}{dx} = \sin x \cdot \sin 2x \cdot \sin 4x \cdot \sin 8x \left( \cot x + 2 \cot 2x + 4 \cot 4x + 8 \cot 8x \right) \]

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