find the derivative.\n\frac{d}{dt}7(t^{2}+9t)^{-5}\n\n\frac{d}{dt}7(t^{2}+9t)^{-5}=□

Explanation:

Step1: Factor out the constant

Since the derivative of a constant - multiple of a function is the constant times the derivative of the function, and we have $y = 7(t^{2}+9t)^{-5}$, then $\frac{d}{dt}7(t^{2}+9t)^{-5}=7\frac{d}{dt}(t^{2}+9t)^{-5}$.

Step2: Apply the chain - rule

The chain - rule states that if $y = u^{n}$ and $u = g(t)$, then $\frac{dy}{dt}=\frac{dy}{du}\cdot\frac{du}{dt}$. Let $u=t^{2}+9t$ and $n = - 5$. First, find $\frac{dy}{du}$: $\frac{d}{du}u^{-5}=-5u^{-6}$. Second, find $\frac{du}{dt}$: $\frac{d}{dt}(t^{2}+9t)=2t + 9$.

Step3: Calculate the derivative

By the chain - rule, $\frac{d}{dt}(t^{2}+9t)^{-5}=-5(t^{2}+9t)^{-6}\cdot(2t + 9)$. Then $7\frac{d}{dt}(t^{2}+9t)^{-5}=7\times(-5)(t^{2}+9t)^{-6}\cdot(2t + 9)$.

Step4: Simplify the result

$7\times(-5)(t^{2}+9t)^{-6}\cdot(2t + 9)=-35(2t + 9)(t^{2}+9t)^{-6}=-\frac{35(2t + 9)}{(t^{2}+9t)^{6}}$.

Answer:

$-\frac{35(2t + 9)}{(t^{2}+9t)^{6}}$