Question

find the derivative of the function: $h(t)=\frac{t^{5}+8t^{4}-5t^{2}}{7t^{2}}$
$\frac{3}{7}t^{3}+\frac{16}{7}t^{2}-\frac{5}{7}$
$\frac{3}{7}t^{2}+\frac{16}{7}t-\frac{5}{7}$
no correct answer choice given.
$\frac{5}{7}t^{3}+\frac{10}{7}t^{2}$
$\frac{5}{7}t^{2}+\frac{10}{7}t$

Explanation:

Step1: Simplify the function

First, divide each term in the numerator by $7t^{2}$:
\[

$$\begin{align*} h(t)&=\frac{t^{5}+8t^{4}-5t^{2}}{7t^{2}}\\ &=\frac{t^{5}}{7t^{2}}+\frac{8t^{4}}{7t^{2}}-\frac{5t^{2}}{7t^{2}}\\ &=\frac{1}{7}t^{3}+\frac{8}{7}t^{2}-\frac{5}{7} \end{align*}$$

\]

Step2: Apply the power - rule for differentiation

The power - rule states that if $y = ax^{n}$, then $y^\prime=anx^{n - 1}$.
For $y=\frac{1}{7}t^{3}$, the derivative $y^\prime=\frac{1}{7}\times3t^{2}=\frac{3}{7}t^{2}$.
For $y = \frac{8}{7}t^{2}$, the derivative $y^\prime=\frac{8}{7}\times2t=\frac{16}{7}t$.
For $y=-\frac{5}{7}$ (a constant), the derivative $y^\prime = 0$.
So, $h^\prime(t)=\frac{3}{7}t^{2}+\frac{16}{7}t$.

Answer:

$\frac{3}{7}t^{2}+\frac{16}{7}t$