To differentiate \( y = x^{\sin{x}} \), we can use logarithmic differentiation. Here's the process:

1. **Take the natural logarithm of both sides** to make the exponent easier to handle:
   \[
   \ln{y} = \ln{(x^{\sin{x}})}
   \]
   Using the property of logarithms \( \ln{(a^b)} = b\ln{a} \), this becomes:
   \[
   \ln{y} = \sin{x} \cdot \ln{x}
   \]

2. **Differentiate both sides** with respect to \( x \) (using implicit differentiation on the left and the product rule on the right):
   \[
   \frac{1}{y} \cdot \frac{dy}{dx} = \cos{x} \cdot \ln{x} + \sin{x} \cdot \frac{1}{x}
   \]

3. **Solve for \( \frac{dy}{dx} \)** by multiplying both sides by \( y \):
   \[
   \frac{dy}{dx} = y \left( \cos{x} \cdot \ln{x} + \frac{\sin{x}}{x} \right)
   \]

4. **Substitute back \( y = x^{\sin{x}} \)**:
   \[
   \frac{dy}{dx} = x^{\sin{x}} \left( \cos{x} \cdot \ln{x} + \frac{\sin{x}}{x} \right)
   \]

Thus, the derivative of \( y = x^{\sin{x}} \) is:
\[
\frac{dy}{dx} = x^{\sin{x}} \left( \cos{x} \cdot \ln{x} + \frac{\sin{x}}{x} \right)
\]

Would you like any additional details or further explanation?

Here are 5 related questions:
1. How would you find the second derivative of \( y = x^{\sin{x}} \)?
2. Can logarithmic differentiation be used for other types of functions? If so, how?
3. What is the domain of \( x^{\sin{x}} \) for real values of \( x \)?
4. How does the derivative behave near \( x = 0 \)?
5. How can we extend this to functions of the form \( x^{f(x)} \)?

**Tip**: When faced with an exponent involving a variable, logarithmic differentiation is a powerful tool to simplify differentiation.

Discover how to differentiate the function y = x^sinx using logarithmic differentiation. This advanced calculus problem involves key concepts like the product rule and chain rule, making it suitable for students in Grades 11-12.

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