https://file.aiquickdraw.com/mathbot/file/1729375319350bp6v80hm.jpg

To find the equation of the tangent line to the graph of \( y = \ln(5 - x^2) \) at \( x = 2 \), we can follow these steps:

### Step 1: Differentiate the function
The general form of the equation for the tangent line is:
\[
y - y_1 = m(x - x_1)
\]
where \( m \) is the slope of the tangent line, and \( (x_1, y_1) \) is the point of tangency. To find \( m \), we need the derivative of \( y \).

The function given is:
\[
y = \ln(5 - x^2)
\]

Using the chain rule for derivatives, we get:
\[
\frac{dy}{dx} = \frac{1}{5 - x^2} \cdot \frac{d}{dx}(5 - x^2)
\]
The derivative of \( 5 - x^2 \) is \( -2x \), so:
\[
\frac{dy}{dx} = \frac{-2x}{5 - x^2}
\]

### Step 2: Evaluate the derivative at \( x = 2 \)
Now, substitute \( x = 2 \) into the derivative to find the slope \( m \):
\[
\frac{dy}{dx}\bigg|_{x = 2} = \frac{-2(2)}{5 - 2^2} = \frac{-4}{5 - 4} = \frac{-4}{1} = -4
\]
Thus, the slope of the tangent line is \( m = -4 \).

### Step 3: Find the point of tangency
Next, find the \( y \)-coordinate of the point of tangency by substituting \( x = 2 \) into the original function:
\[
y = \ln(5 - 2^2) = \ln(5 - 4) = \ln(1) = 0
\]
So, the point of tangency is \( (2, 0) \).

### Step 4: Write the equation of the tangent line
Using the point-slope form of the equation of a line:
\[
y - y_1 = m(x - x_1)
\]
Substitute \( y_1 = 0 \), \( m = -4 \), and \( x_1 = 2 \):
\[
y - 0 = -4(x - 2)
\]
\[
y = -4(x - 2)
\]

### Final Answer:
The equation of the tangent line is:
\[
y = -4(x - 2)
\]

Would you like further details or explanations? Here are some related questions for practice:

1. What is the significance of the derivative in determining the slope of a tangent line?
2. How does the chain rule apply when differentiating composite functions like \( \ln(5 - x^2) \)?
3. What is the meaning of the point-slope form of a line equation?
4. Can you find the tangent line at a different point, say \( x = 1 \), for the same function?
5. What happens to the tangent line if \( x = 0 \)? Does the tangent exist there?

**Tip**: Always check that the argument inside a logarithmic function is positive before evaluating!

Find the equation of the tangent line to the graph of y = ln(5 - x^2) at x = 2.

Learn how to find the equation of the tangent line to y = ln(5 - x^2) at x = 2. This solution involves logarithmic differentiation, the chain rule, and finding the slope of the tangent line. Suitable for advanced high school or college-level calculus students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation