find the derivative.\n$\frac{d}{dt}7(t^{2}+9t)^{-3}$\n$\frac{d}{dt}7(t^{2}+9t)^{-3}=square$

Explanation:

Step1: Factor out the constant

Since $\frac{d}{dt}(cf(t)) = c\frac{d}{dt}f(t)$ for a constant $c$, we have $\frac{d}{dt}7(t^{2}+9t)^{-3}=7\frac{d}{dt}(t^{2}+9t)^{-3}$.

Step2: Apply the chain - rule

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

Step3: Multiply by the constant

$7\frac{d}{dt}(t^{2}+9t)^{-3}=7\times(-3)(t^{2}+9t)^{-4}\cdot(2t + 9)=-21(2t + 9)(t^{2}+9t)^{-4}$.

Answer:

$-21(2t + 9)(t^{2}+9t)^{-4}$