Question

find h(t) if h(t)=14.6 + 0.6t - 1.8t^2.
h(t)=□

Explanation:

Step1: Recall derivative rules

The derivative of a constant $C$ is $0$, i.e., $\frac{d}{dt}(C) = 0$; the derivative of $at$ with respect to $t$ is $a$ ($\frac{d}{dt}(at)=a$), and the derivative of $at^{n}$ with respect to $t$ is $nat^{n - 1}$ ($\frac{d}{dt}(at^{n})=nat^{n-1}$).

Step2: Differentiate each term

For the constant term $14.6$, $\frac{d}{dt}(14.6)=0$. For the term $0.6t$, $\frac{d}{dt}(0.6t)=0.6$. For the term $- 1.8t^{2}$, using the power - rule $\frac{d}{dt}(-1.8t^{2})=-1.8\times2t=-3.6t$.

Step3: Combine the results

$h^{\prime}(t)=\frac{d}{dt}(14.6)+\frac{d}{dt}(0.6t)-\frac{d}{dt}(1.8t^{2})=0 + 0.6-3.6t$.

Answer:

$0.6 - 3.6t$