find the derivative.\n\frac{d}{dt}\frac{(9t - 4)^7}{t + 8}\n\n\frac{d}{dt}\frac{(9t - 4)^7}{t + 8}=square

Explanation:

Step1: Apply quotient - rule

The quotient - rule states that if $y=\frac{u}{v}$, then $y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}$. Here, $u=(9t - 4)^{7}$ and $v=t + 8$.

Step2: Find $u^\prime$ using chain - rule

Let $w = 9t-4$, so $u = w^{7}$. By the chain - rule $\frac{du}{dt}=\frac{du}{dw}\cdot\frac{dw}{dt}$. $\frac{du}{dw}=7w^{6}=7(9t - 4)^{6}$ and $\frac{dw}{dt}=9$. So, $u^\prime=63(9t - 4)^{6}$.

Step3: Find $v^\prime$

Since $v=t + 8$, then $v^\prime = 1$.

Step4: Substitute into quotient - rule

$\frac{d}{dt}\frac{(9t - 4)^{7}}{t + 8}=\frac{63(9t - 4)^{6}(t + 8)-(9t - 4)^{7}\times1}{(t + 8)^{2}}$. Factor out $(9t - 4)^{6}$ from the numerator: [ \begin{align*} &=\frac{(9t - 4)^{6}[63(t + 8)-(9t - 4)]}{(t + 8)^{2}}\ &=\frac{(9t - 4)^{6}(63t+504 - 9t + 4)}{(t + 8)^{2}}\ &=\frac{(9t - 4)^{6}(54t+508)}{(t + 8)^{2}}\ &=\frac{2(9t - 4)^{6}(27t + 254)}{(t + 8)^{2}} \end{align*} ]

Answer:

$\frac{2(9t - 4)^{6}(27t + 254)}{(t + 8)^{2}}$