To differentiate the function \( 5x^3 \cdot e^{\ln(\sin(\arcsin((x^2+1)/x)))} \) with respect to \( x \), let's go through the process step by step.

### Step 1: Simplify the Exponential Term
The expression \( e^{\ln(\sin(\arcsin((x^2+1)/x)))} \) can be simplified using the property \( e^{\ln(f(x))} = f(x) \). So, we have:
\[
e^{\ln(\sin(\arcsin((x^2+1)/x)))} = \sin(\arcsin((x^2+1)/x))
\]
Since \( \sin(\arcsin(y)) = y \), the expression further simplifies to:
\[
\sin(\arcsin((x^2+1)/x)) = \frac{x^2+1}{x}
\]
Thus, the function now is:
\[
f(x) = 5x^3 \cdot \frac{x^2+1}{x}
\]

### Step 2: Simplify the Function
Simplifying the expression gives:
\[
f(x) = 5x^3 \cdot \left(\frac{x^2+1}{x}\right) = 5x^3 \cdot \left(x + \frac{1}{x}\right) = 5x^4 + 5x^2
\]

### Step 3: Differentiate the Function
Now, differentiate \( f(x) = 5x^4 + 5x^2 \) with respect to \( x \):
\[
f'(x) = \frac{d}{dx}(5x^4) + \frac{d}{dx}(5x^2)
\]

Applying the power rule \( \frac{d}{dx}(x^n) = nx^{n-1} \) to each term:
\[
f'(x) = 20x^3 + 10x
\]

So, the derivative of the given function is:
\[
\boxed{20x^3 + 10x}
\]

Would you like more details or have any questions? Here are five related questions:

1. How do you apply the power rule in differentiation?
2. What is the chain rule, and when is it used?
3. How does one differentiate exponential functions involving natural logarithms?
4. Can you explain the relationship between inverse trigonometric functions and their derivatives?
5. How would you find the derivative of a product of three functions?

**Tip:** Simplifying a function before differentiating can often make the process easier and reduce the likelihood of errors.

Learn how to differentiate the function 5x^3 * e^ln(sin(arcsin((x^2+1)/x))) step-by-step. This problem involves differentiation, exponential functions, natural logarithms, and inverse trigonometric functions. Suitable for advanced high school students.

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