consider the following function.\nf(t) = (\\frac{1}{9t + 1})^3\nsimplify by rewriting f(t) using a negative…

consider the following function.\nf(t) = (\\frac{1}{9t + 1})^3\nsimplify by rewriting f(t) using a negative exponent and no fractions.\nf(t) =\nfind the derivative of the simplified function.\nf(t) =

consider the following function.\nf(t) = (\\frac{1}{9t + 1})^3\nsimplify by rewriting f(t) using a negative exponent and no fractions.\nf(t) =\nfind the derivative of the simplified function.\nf(t) =

Answer

Explanation:

Step1: Rewrite using negative exponent

According to the rule $\frac{1}{a}=a^{- 1}$, we have $F(t)=\left(\frac{1}{9t + 1}\right)^3=(9t + 1)^{-3}$.

Step2: Find the derivative

Use the chain - rule. If $y = u^{-3}$ and $u=9t + 1$, then $\frac{dy}{du}=-3u^{-4}$ and $\frac{du}{dt}=9$. By the chain - rule $\frac{dy}{dt}=\frac{dy}{du}\cdot\frac{du}{dt}$. Substitute $u = 9t+1$ back in: $F^{\prime}(t)=-3(9t + 1)^{-4}\cdot9=-27(9t + 1)^{-4}$.

Answer:

$F(t)=(9t + 1)^{-3}$ $F^{\prime}(t)=-27(9t + 1)^{-4}$