Question

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find the derivative of the function.
f(t) = 1/2 (9t^2 + t)^(-3)
f(t) =
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Explanation:

Step1: Apply the constant - multiple rule

The constant - multiple rule states that if $y = cf(x)$, then $y'=cf'(x)$. Here $c=\frac{1}{2}$ and $y = f(t)=\frac{1}{2}(9t^{2}+t)^{-3}$, so $f'(t)=\frac{1}{2}\frac{d}{dt}(9t^{2}+t)^{-3}$.

Step2: Apply the chain rule

Let $u = 9t^{2}+t$, then $y = u^{-3}$. The chain rule states that $\frac{dy}{dt}=\frac{dy}{du}\cdot\frac{du}{dt}$. First, find $\frac{dy}{du}$: $\frac{dy}{du}=- 3u^{-4}=-3(9t^{2}+t)^{-4}$. Second, find $\frac{du}{dt}$: $\frac{du}{dt}=\frac{d}{dt}(9t^{2}+t)=18t + 1$.

Step3: Calculate the derivative

By the chain - rule $\frac{d}{dt}(9t^{2}+t)^{-3}=\frac{dy}{du}\cdot\frac{du}{dt}=-3(9t^{2}+t)^{-4}(18t + 1)$. Then $f'(t)=\frac{1}{2}\times(-3(9t^{2}+t)^{-4}(18t + 1))=-\frac{3(18t + 1)}{2(9t^{2}+t)^{4}}$.

Answer:

$-\frac{3(18t + 1)}{2(9t^{2}+t)^{4}}$