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!