5. - / 1 points find the derivative of the function. $f(t)=left(\frac{1}{8t + 1}\right)^3$ $f(t)=$

Explanation:

Step1: Rewrite the function

Rewrite $F(t)=\left(\frac{1}{8t + 1}\right)^3=(8t + 1)^{-3}$.

Step2: Apply the chain - rule

The chain - rule states that if $y = u^n$ and $u = g(t)$, then $\frac{dy}{dt}=n\cdot u^{n - 1}\cdot g^\prime(t)$. Here $n=-3$ and $u = 8t+1$, and $g(t)=8t + 1$, so $g^\prime(t)=8$.

Step3: Calculate the derivative

$F^\prime(t)=-3(8t + 1)^{-3 - 1}\cdot8$. Simplify the expression: $F^\prime(t)=-24(8t + 1)^{-4}=-\frac{24}{(8t + 1)^4}$.

Answer:

$-\frac{24}{(8t + 1)^4}$