https://r2.erweima.ai/mathbot/image/1725569707438cn7q9lu3.jpg

The problem asks for the derivative of \( y = \frac{x^x}{1+x^x} \), based on the image.

Let's find the derivative of \( y \) with respect to \( x \).

### Step-by-step:

1. **Rewrite \( x^x \)**: 
   Since \( x^x = e^{x \ln x} \), this will be helpful when differentiating.

2. **Use the quotient rule**:
   The quotient rule for a function \( y = \frac{u(x)}{v(x)} \) is:
   \[
   y' = \frac{u'(x)v(x) - u(x)v'(x)}{v(x)^2}
   \]
   where \( u(x) = x^x \) and \( v(x) = 1 + x^x \).

3. **Differentiate \( u(x) = x^x \)**:
   \[
   u'(x) = \frac{d}{dx}\left(e^{x \ln x}\right) = e^{x \ln x} \left( \ln x + 1 \right) = x^x (\ln x + 1)
   \]

4. **Differentiate \( v(x) = 1 + x^x \)**:
   \[
   v'(x) = 0 + \frac{d}{dx}(x^x) = x^x (\ln x + 1)
   \]

5. **Apply the quotient rule**:
   \[
   y' = \frac{x^x (\ln x + 1)(1 + x^x) - x^x (x^x (\ln x + 1))}{(1 + x^x)^2}
   \]
   Simplify the numerator:
   \[
   = \frac{x^x (\ln x + 1) + x^{2x} (\ln x + 1) - x^{2x} (\ln x + 1)}{(1 + x^x)^2}
   \]
   The terms involving \( x^{2x} (\ln x + 1) \) cancel out, leaving:
   \[
   y' = \frac{x^x (\ln x + 1)}{(1 + x^x)^2}
   \]

This matches option **(c)** in the image. 

So, the correct answer is:
\[
\boxed{c \: \frac{x^x (\ln x + 1)}{(1 + x^x)^2}}
\]

Do you have any questions or would you like more details?

### Related questions:
1. How does the quotient rule apply to other functions with similar forms?
2. What is the derivative of \( x^x \) and how is it computed in general?
3. Can this derivative be simplified further if \( x \) has specific values?
4. What are other methods to approach differentiating products like \( x^x \)?
5. How does logarithmic differentiation simplify handling powers involving \( x \)?

**Tip**: When dealing with expressions involving powers like \( x^x \), logarithmic differentiation is a useful technique to simplify calculations.

Learn how to differentiate the function y = x^x / (1 + x^x) step-by-step using the quotient rule and logarithmic differentiation. This problem covers advanced calculus concepts, making it ideal for college-level students.

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