Question

1. evaluate $\frac{d}{dt}(3e^{-8t})$. $\frac{d}{dt}(3e^{-8t})=$

Explanation:

Step1: Apply constant - multiple rule

The constant - multiple rule of differentiation states that if $y = cf(x)$, then $\frac{dy}{dx}=c\frac{df(x)}{dx}$, where $c$ is a constant. Here $c = 3$ and $f(t)=e^{-8t}$, so $\frac{d}{dt}(3e^{-8t})=3\frac{d}{dt}(e^{-8t})$.

Step2: Apply chain - rule for $e^{-8t}$

The chain - rule states that if $y = e^{u}$ and $u=-8t$, then $\frac{dy}{dt}=\frac{dy}{du}\cdot\frac{du}{dt}$. The derivative of $y = e^{u}$ with respect to $u$ is $\frac{dy}{du}=e^{u}$, and the derivative of $u=-8t$ with respect to $t$ is $\frac{du}{dt}=-8$. So $\frac{d}{dt}(e^{-8t})=e^{-8t}\cdot(-8)$.

Step3: Combine results

Substitute $\frac{d}{dt}(e^{-8t}) = - 8e^{-8t}$ into $3\frac{d}{dt}(e^{-8t})$. We get $3\times(-8e^{-8t})=-24e^{-8t}$.

Answer:

$-24e^{-8t}$