Answer
The value of the derivative is: $\lim \limits_{h \to 0}(-0.6t-0.3h)$, which equals to $-0.6t$
$R'(2)=-0.6\times 2=-1.2$
Work Step by Step
The algebraic definition of a derivative can be written as:
$R'(t)=\lim \limits_{h \to 0} \frac{R(t+h)-R(t)}{h}$
By substituting the given function we get:
$\frac{R(t+h)-R(t)}{h}=\frac{-0.3(t+h)^2-(-0.3t^2)}{h}=\frac{-0.3t^2-0.6th-0.3h^2+0.3t^2}{h}=\frac{-0.6th-0.3h^2}{h}=-0.6t-0.3h$
The value of the derivative is: $\lim \limits_{h \to 0}(-0.6t-0.3h)$, which equals to $-0.6t$
$R'(2)=-0.6\times 2=-1.2$
