Question

Calculus early transcendental functions. Find the derivative using the quotient rule

Answer 1

$f^{\prime}(t)=\frac{-14}{t^3}$

Explanation:

$\begin{gathered} f(t)\text{ = }\frac{7}{t^2}\text{ + }\frac{4}{t^5} \\ finding\text{ the LCM:} \\ f(t)\text{ = }\frac{7t^3+\text{ 4}}{t^5} \end{gathered}$

Using quotient rule:

$\begin{gathered} f^{\prime}(t)\text{ = }\frac{\text{v}\frac{du}{d\text{ t}}\text{ - u }\frac{dv}{d\text{ t}}}{v^2} \\ u=7t^3+4,du/dt=3(7)t^2=21t^2 \\ v=t^5,dv/dt=5t^4 \\ \\ f^{\prime}(t)\text{ = }\frac{d}{dt}(\frac{7t^3+\text{ 4}}{t^5}) \end{gathered}$$\begin{gathered} f^{\prime}(t)\text{ = }\frac{t^5(21t^2)-7t^3(5t^4)}{(t^5)^2} \\ f^{\prime}(t)\text{ = }\frac{21t^7-35t^7}{t^{10}^{}} \end{gathered}$$\begin{gathered} f^{\prime}(t)\text{ = }\frac{t^7(21-35)}{t^{10}} \\ f^{\prime}(t)\text{ = }t^7\times t^{-10}\times(21-35) \\ f^{\prime}(t)=t^{-3}(21-35)\text{ }=t^{-3}(-14) \\ f^{\prime}(t)=\frac{-14}{t^3} \end{gathered}$