We are tasked with using logarithmic differentiation to find the derivative of the function:

$g(x) = \left( \sin(2x) \right)^x$

Step-by-Step Solution

1. Take the natural logarithm (ln) of both sides:

   $\ln(g(x)) = \ln\left( (\sin(2x))^x \right)$

   Using the logarithmic property $\ln(a^b) = b\ln(a)$, this simplifies to:

   $\ln(g(x)) = x \ln(\sin(2x))$

2. Differentiate both sides with respect to $x$ using implicit differentiation. Keep in mind that the derivative of $\ln(g(x))$ is $\frac{g'(x)}{g(x)}$.

   $\frac{d}{dx} \left( \ln(g(x)) \right) = \frac{d}{dx} \left( x \ln(\sin(2x)) \right)$

   The left side becomes:

   $\frac{g'(x)}{g(x)}$

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

   This simplifies to:

   $\ln(\sin(2x)) + x \cdot \frac{1}{\sin(2x)} \cdot \frac{d}{dx}(\sin(2x))$

3. Differentiate $\sin(2x)$:

   $\frac{d}{dx}(\sin(2x)) = 2\cos(2x)$

   So the expression becomes:

   $\ln(\sin(2x)) + x \cdot \frac{2\cos(2x)}{\sin(2x)}$

4. Substitute this result back into the equation:

   $\frac{g'(x)}{g(x)} = \ln(\sin(2x)) + 2x \cdot \cot(2x)$

5. Solve for $g'(x)$:

   Multiply both sides by $g(x) = (\sin(2x))^x$:

   $g'(x) = (\sin(2x))^x \left( \ln(\sin(2x)) + 2x \cdot \cot(2x) \right)$

Thus, the derivative of $g(x) = (\sin(2x))^x$ is:

$g'(x) = (\sin(2x))^x \left( \ln(\sin(2x)) + 2x \cdot \cot(2x) \right)$

Would you like further details or have any other questions?

Here are 5 related questions to expand on this topic:

1. How is logarithmic differentiation useful when differentiating functions of the form $f(x)^g(x)$?
2. What are some common logarithmic properties that simplify complex derivatives?
3. How would the derivative change if the base of the power was another trigonometric function, such as $\cos(2x)$?
4. Can logarithmic differentiation be applied to implicit functions? If yes, how?
5. What are the differences between logarithmic differentiation and regular differentiation in terms of steps?

Tip: Logarithmic differentiation is particularly effective when both the base and the exponent are functions of $x$, as it simplifies the application of the product and chain rules.