To find the derivative of the function:

\[
f(x) = \frac{e^{5x} \sin^{-1}(x)}{\ln(2x)}
\]

we'll use the quotient rule and the product rule, as the function is a quotient where the numerator itself involves a product of two functions. The quotient rule states:

\[
\left( \frac{u}{v} \right)' = \frac{v u' - u v'}{v^2}
\]

where \( u = e^{5x} \sin^{-1}(x) \) and \( v = \ln(2x) \).

### Step 1: Differentiate \( u = e^{5x} \sin^{-1}(x) \) 

Using the product rule for \( u \), where:
- \( u_1 = e^{5x} \) with \( u_1' = 5 e^{5x} \)
- \( u_2 = \sin^{-1}(x) \) with \( u_2' = \frac{1}{\sqrt{1 - x^2}} \)

Then,
\[
u' = u_1' u_2 + u_1 u_2' = (5 e^{5x}) \sin^{-1}(x) + e^{5x} \cdot \frac{1}{\sqrt{1 - x^2}}
\]

### Step 2: Differentiate \( v = \ln(2x) \)

Using the chain rule for \( v \):
\[
v' = \frac{1}{2x} \cdot 2 = \frac{1}{x}
\]

### Step 3: Apply the Quotient Rule

Substitute \( u \), \( v \), \( u' \), and \( v' \) into the quotient rule:

\[
f'(x) = \frac{\ln(2x) \left[ (5 e^{5x}) \sin^{-1}(x) + \frac{e^{5x}}{\sqrt{1 - x^2}} \right] - e^{5x} \sin^{-1}(x) \cdot \frac{1}{x}}{(\ln(2x))^2}
\]

This is the derivative \( f'(x) \) of the given function. 

Would you like a more detailed breakdown or simplification? 

Here are some follow-up questions to test your understanding:
1. What is the derivative of \( \sin^{-1}(x) \) and why does it involve \( \sqrt{1 - x^2} \)?
2. How would the process change if \( e^{5x} \) was replaced with \( e^{5x^2} \)?
3. What is the result if you apply the product rule to functions involving trigonometric expressions?
4. How does the chain rule apply to \( \ln(2x) \)?
5. Can you identify scenarios where the quotient rule might not apply directly?

**Tip:** Remember, breaking down each function separately is crucial in complex derivatives.

Learn to differentiate (e^(5x) * sin^-1(x)) / ln(2x) using the quotient and product rules. This solution covers complex differentiation techniques, including the chain rule and inverse trigonometric derivatives. Suitable for advanced calculus students in Grades 11-12.

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