Question

question 2
find the derivative of the function.
h(t)=\frac{t^{9}+7t^{5}+9t^{2}}{8t^{2}}
\\(\frac{7}{8}t^{6}+\frac{21}{8}t^{2}+\frac{9}{8}\\)
\\(\frac{7}{8}t^{7}+\frac{21}{8}t^{3}+\frac{9}{8}\\)
\\(\frac{1}{8}t^{7}+\frac{21}{8}t^{3}\\)
\\(\frac{1}{8}t^{6}+\frac{21}{8}t^{2}\\)

Explanation:

Step1: Simplify the function

First, divide each term in the numerator by $8t^{2}$:
\[h(t)=\frac{t^{9}}{8t^{2}}+\frac{7t^{5}}{8t^{2}}+\frac{9t^{2}}{8t^{2}}=\frac{1}{8}t^{7}+\frac{7}{8}t^{3}+\frac{9}{8}\]

Step2: Apply the power - rule for differentiation

The power - rule states that if $y = ax^{n}$, then $y^\prime=anx^{n - 1}$.
For the first term $\frac{1}{8}t^{7}$, its derivative is $\frac{1}{8}\times7t^{7 - 1}=\frac{7}{8}t^{6}$.
For the second term $\frac{7}{8}t^{3}$, its derivative is $\frac{7}{8}\times3t^{3 - 1}=\frac{21}{8}t^{2}$.
For the third term $\frac{9}{8}$ (a constant), its derivative is 0.
So, $h^\prime(t)=\frac{7}{8}t^{6}+\frac{21}{8}t^{2}$.

Answer:

$\frac{7}{8}t^{6}+\frac{21}{8}t^{2}$