Question

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

Explanation:

Step1: Use constant - multiple rule

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

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

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

Step3: Combine results

Substitute the result of $\frac{d}{dt}(e^{-8t})$ into the expression from Step 1. We have $3\frac{d}{dt}(e^{-8t})=3\times(-8)e^{-8t}=- 24e^{-8t}$.

Answer:

$-24e^{-8t}$