18. what is the slope of the line tangent to the graph of $\frac{x^{2}-2}{x^{2}+1}$ when $x = 1$?
18. what is the slope of the line tangent to the graph of $\frac{x^{2}-2}{x^{2}+1}$ when $x = 1$?
Answer
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: [ \begin{align*} (2x)(x^{2}+1)-(x^{2}-2)(2x)&=2x^{3}+2x-(2x^{3}-4x)\ &=2x^{3}+2x - 2x^{3}+4x\ &=6x \end{align*} ] 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})