## Calculus: Early Transcendentals (2nd Edition)

$y' = 1$
Quotient Rule: $y=\frac{x^2-a^2}{x-a}$ $y' = \frac{(2x)(x-a)-(x^2-a^2)(1)}{(x-a)^2} = 1$ Division then Power Rule: $y=\frac{x^2-a^2}{x-a} = \frac{(x-a)(x+a)}{x-a} = x+a$ $y' = 1$