## Calculus: Early Transcendentals (2nd Edition)

The value of the derivative is $$\left.\frac{dy}{dt}\right|_{t=2}=-3.$$
Using the definition of the derivative and putting $t=2$ we have $$\left.\frac{dy}{dt}\right|_{t=2}=\lim_{h\to0}\frac{f(2+h)-f(2)}{h}=\lim_{h\to0}\frac{(2+h)-(2+h)^2-(2-2^2)}{h}=\lim_{h\to0}\frac{2+h-(2^2+h^2+4h)+2}{h}=\lim_{h\to0}\frac{-2-h^2-3h+2}{h}=\lim_{h\to0}\frac{-h^2-3h}{h}=\lim_{h\to0}(-h-3)=-0-3=-3.$$