Question

1. what is the slope of the line tangent to the graph of $\frac{x^{2}-2}{x^{2}+1}$ when $x = 1$?

Explanation:

Step1: Use the quotient rule

The quotient rule states that if \(y = \frac{u}{v}\), then \(y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}\). Let \(u=x^{2}-2\), so \(u^\prime = 2x\), and \(v=x^{2}+1\), so \(v^\prime=2x\).

$$y^\prime=\frac{(2x)(x^{2}+1)-(x^{2}-2)(2x)}{(x^{2}+1)^{2}}$$

Step2: Simplify the numerator

Expand the numerator:

$$
LATEXBLOCK0
$$

So \(y^\prime=\frac{6x}{(x^{2}+1)^{2}}\)

Step3: Substitute \(x = 1\)

When \(x = 1\), \(y^\prime=\frac{6\times1}{(1^{2}+1)^{2}}=\frac{6}{4}=\frac{3}{2}\)

Answer:

\(\frac{3}{2}\)