find the derivative.\n\\(\\frac{d}{dt}\\frac{(3t - 4)^3}{t + 5}\\)\n\\(\\frac{d}{dt}\\frac{(3t - 4)^3}{t +…

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, $u=(3t - 4)^{3}$ and $v=t + 5$.

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

Let $w = 3t-4$, so $u = w^{3}$. By the chain - rule $\frac{du}{dt}=\frac{du}{dw}\cdot\frac{dw}{dt}$. $\frac{du}{dw}=3w^{2}=3(3t - 4)^{2}$ and $\frac{dw}{dt}=3$. Then $u^\prime=9(3t - 4)^{2}$. Also, $v^\prime = 1$.

Step3: Substitute into quotient - rule

$\frac{d}{dt}\frac{(3t - 4)^{3}}{t + 5}=\frac{9(3t - 4)^{2}(t + 5)-(3t - 4)^{3}\times1}{(t + 5)^{2}}$. Factor out $(3t - 4)^{2}$: $\frac{(3t - 4)^{2}[9(t + 5)-(3t - 4)]}{(t + 5)^{2}}$. Expand the numerator: $\frac{(3t - 4)^{2}(9t+45 - 3t + 4)}{(t + 5)^{2}}=\frac{(3t - 4)^{2}(6t + 49)}{(t + 5)^{2}}$.

Answer:

$\frac{(3t - 4)^{2}(6t + 49)}{(t + 5)^{2}}$