Find an equation of the line tangent to the following equation at point xequals=1.
y equals x Superscript cotangent xy=xcotx
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Part 1
The equation of the tangent line is y equalsy=.
​(Type an equation using x and y as the variables. Type an exact​ answer.)

To find the equation of the tangent line, we follow these steps:

### 1. Implicit Differentiation of \( y = x \cot(x) \)

The equation given is:
\[
y = x \cot(x)
\]

First, differentiate both sides with respect to \( x \).

\[
\frac{d}{dx}(y) = \frac{d}{dx} \left( x \cot(x) \right)
\]

Using the product rule for differentiation on the right-hand side:
\[
\frac{d}{dx} \left( x \cot(x) \right) = \cot(x) + x \cdot \frac{d}{dx} \left( \cot(x) \right)
\]
The derivative of \( \cot(x) \) is \( -\csc^2(x) \), so:
\[
\frac{d}{dx} \left( x \cot(x) \right) = \cot(x) - x \csc^2(x)
\]

Thus, the derivative becomes:
\[
\frac{dy}{dx} = \cot(x) - x \csc^2(x)
\]

### 2. Evaluate the Derivative at \( x = 1 \)

Next, we evaluate \( \frac{dy}{dx} \) at \( x = 1 \):

\[
\frac{dy}{dx} \Bigg|_{x=1} = \cot(1) - 1 \cdot \csc^2(1)
\]

Since \( \cot(1) \) and \( \csc(1) \) are transcendental values, we'll leave them in exact form. The slope \( m \) of the tangent line at \( x = 1 \) is:
\[
m = \cot(1) - \csc^2(1)
\]

### 3. Find the Point on the Curve at \( x = 1 \)

To find the point on the curve, substitute \( x = 1 \) into the original equation \( y = x \cot(x) \):

\[
y = 1 \cdot \cot(1)
\]

Thus, the point is \( (1, \cot(1)) \).

### 4. Equation of the Tangent Line

The equation of the tangent line is given by:
\[
y - y_1 = m(x - x_1)
\]
Substituting \( x_1 = 1 \), \( y_1 = \cot(1) \), and \( m = \cot(1) - \csc^2(1) \):

\[
y - \cot(1) = \left( \cot(1) - \csc^2(1) \right)(x - 1)
\]

This is the equation of the tangent line in exact form.

### Final Equation
\[
y = \left( \cot(1) - \csc^2(1) \right)(x - 1) + \cot(1)
\]

Would you like further clarification on any step or additional details?

Here are 5 related questions to further explore this concept:
1. How do you perform implicit differentiation when both \( x \) and \( y \) are mixed?
2. What are the derivative rules for trigonometric functions?
3. How would the equation of the tangent line change if the point of tangency was \( x = 2 \)?
4. What is the geometric interpretation of the derivative at a point?
5. Can you find the normal line at \( x = 1 \) for the same curve?

**Tip:** The product rule is crucial when differentiating expressions involving both a function of \( x \) and a trigonometric function.

Find an equation of the line tangent to the following equation at point x = 1. y = x * cot(x)

Learn how to find the equation of the tangent line for y = x * cot(x) at x = 1 using implicit differentiation. This solution explores key calculus concepts, including the product rule and the derivatives of trigonometric functions. Suitable for high school or college calculus students.

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