The equation in the image is:

$\log\left(x^2 e^{-3x}\right) \sin(2x)$

To differentiate this expression:

Let: $y = \log\left(x^2 e^{-3x}\right) \sin(2x)$

We need to use both the product rule and chain rule of differentiation. The product rule states: $\frac{d}{dx}(uv) = u'v + uv'$ where $u = \log\left(x^2 e^{-3x}\right)$ and $v = \sin(2x)$.

Step 1: Differentiate $u = \log\left(x^2 e^{-3x}\right)$

$u = \log\left(x^2 e^{-3x}\right) = \log(x^2) + \log(e^{-3x})$

Using properties of logarithms: $u = 2\log(x) - 3x$

Now differentiate $u$: $u' = \frac{d}{dx} \left( 2\log(x) - 3x \right) = \frac{2}{x} - 3$

Step 2: Differentiate $v = \sin(2x)$

Using the chain rule: $v' = \frac{d}{dx} \sin(2x) = 2 \cos(2x)$

Step 3: Apply the product rule

Now apply the product rule to $y = u \cdot v$: $y' = u'v + uv'$

Substitute the values of $u'$, $v$, and $v'$: $y' = \left( \frac{2}{x} - 3 \right) \sin(2x) + \left( 2 \log(x) - 3x \right) \cdot 2 \cos(2x)$

Thus, the derivative of the given function is: $y' = \left( \frac{2}{x} - 3 \right) \sin(2x) + 2 \left( 2\log(x) - 3x \right) \cos(2x)$

Do you want any more details on the differentiation process or further explanation?

Here are 5 related questions to explore further:

1. How do we use the chain rule in more complex logarithmic differentiation problems?
2. Can we simplify the derivative expression further?
3. What is the importance of logarithmic properties in calculus?
4. How would this expression change if it were $\log\left(e^{3x} / x^2\right)$?
5. What are the real-life applications of logarithmic and trigonometric derivatives?

Tip: Always check for simplifications using logarithmic properties to make differentiation easier.