The function provided in the image is:
$f(x) = (\sin x)^{\cos 2x}$

To find the derivative $f'(x)$, we will use logarithmic differentiation since the exponent is also a function of $x$. Let's go step-by-step:

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Step 1: Take the natural logarithm on both sides.

$\ln f(x) = \ln \left( (\sin x)^{\cos 2x} \right)$ Using the logarithmic identity $\ln(a^b) = b \ln a$: $\ln f(x) = \cos 2x \cdot \ln (\sin x)$

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Step 2: Differentiate both sides with respect to $x$.

We apply the chain rule on the left side and the product rule on the right side.

$\frac{1}{f(x)} f'(x) = \frac{d}{dx} \left( \cos 2x \cdot \ln (\sin x) \right)$

For the right side, using the product rule: $\frac{d}{dx} \left( \cos 2x \cdot \ln (\sin x) \right) = \cos 2x \cdot \frac{1}{\sin x} \cos x + \ln (\sin x) \cdot \left( -2 \sin 2x \right)$

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Step 3: Solve for $f'(x)$.

Now, substitute the expression for $f'(x)$ and multiply by $f(x) = (\sin x)^{\cos 2x}$:

$f'(x) = (\sin x)^{\cos 2x} \cdot \left( \frac{\cos 2x \cos x}{\sin x} - 2 \ln (\sin x) \sin 2x \right)$

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Final Answer:

$f'(x) = (\sin x)^{\cos 2x} \cdot \left( \frac{\cos 2x \cos x}{\sin x} - 2 \ln (\sin x) \sin 2x \right)$

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Would you like me to simplify this further or explain any part of the differentiation?

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Here are 5 follow-up questions related to this problem:

1. How does the chain rule apply in logarithmic differentiation?
2. What are some other functions where logarithmic differentiation is useful?
3. How do you apply the product rule in calculus?
4. What is the domain of the function $f(x) = (\sin x)^{\cos 2x}$?
5. How does logarithmic differentiation compare with implicit differentiation?

Tip: Logarithmic differentiation is often helpful when both the base and the exponent are functions of $x$.