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We are given the function \( f(x) = x^{\cot x} \) and asked to find the equation of the tangent line at \( x = 1 \).

### Steps to find the tangent line:
1. **Find the derivative** of \( f(x) = x^{\cot x} \).
   
   We can use logarithmic differentiation since the exponent \( \cot x \) is a function of \( x \).

   Take the natural logarithm of both sides:
   \[
   \ln(f(x)) = \ln(x^{\cot x}) = \cot x \cdot \ln x
   \]

   Now differentiate both sides with respect to \( x \) using the product and chain rules. Let’s differentiate the right-hand side first:

   \[
   \frac{d}{dx}[\ln(f(x))] = \frac{1}{f(x)} \cdot f'(x)
   \]
   \[
   \frac{d}{dx}[\cot x \ln x] = -\csc^2(x) \cdot \ln x + \frac{\cot x}{x}
   \]

   Multiply both sides by \( f(x) = x^{\cot x} \) to isolate \( f'(x) \):

   \[
   f'(x) = x^{\cot x} \left( -\csc^2(x) \cdot \ln x + \frac{\cot x}{x} \right)
   \]

2. **Evaluate the derivative at \( x = 1 \)**.

   Substituting \( x = 1 \) into \( f(x) = x^{\cot x} \), we get:

   \[
   f(1) = 1^{\cot 1} = 1
   \]

   Now calculate the derivative at \( x = 1 \):

   \[
   f'(1) = 1^{\cot 1} \left( -\csc^2(1) \cdot \ln 1 + \frac{\cot 1}{1} \right)
   \]
   Since \( \ln 1 = 0 \), the first term drops out, and we are left with:
   \[
   f'(1) = \cot 1
   \]

3. **Equation of the tangent line** at \( x = 1 \):

   The equation of the tangent line is given by:
   \[
   y - f(1) = f'(1)(x - 1)
   \]
   Substituting the values \( f(1) = 1 \) and \( f'(1) = \cot 1 \):

   \[
   y - 1 = \cot 1 (x - 1)
   \]
   Simplifying:
   \[
   y = \cot 1 \cdot x - \cot 1 + 1
   \]

Thus, the correct answer is option **C**: \( y = x \cot 1 + 1 - \cot 1 \).

Would you like a deeper explanation of any step or have further questions?

Here are 5 related questions you might find useful:
1. What is logarithmic differentiation, and when is it used?
2. How do we apply the product rule in differentiation?
3. What is the derivative of \( \cot x \), and why is it important here?
4. How do we determine the equation of a tangent line using point-slope form?
5. How can we generalize this process for other functions of the form \( f(x) = x^{g(x)} \)?

**Tip**: When dealing with functions that have a variable in both the base and exponent, logarithmic differentiation simplifies the process.

Find an equation of the line tangent to the following equation at point x = 1: f(x) = x^cot(x). What is the equation of the tangent line at x = 1?

Learn how to find the tangent line to f(x) = x^cot(x) at x = 1 using logarithmic differentiation and point-slope form. This detailed solution includes key concepts such as differentiation, logarithmic differentiation, and the cotangent function. Suitable for advanced high school students.

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