Question

find the first and second derivatives.

$y = \frac{3x^{3}+9}{x}$

$\frac{dy}{dx}=\square$

Explanation:

Step1: Simplify the function

$$y=\frac{3x^{3}+9}{x}=3x^{2}+\frac{9}{x}=3x^{2}+9x^{-1}$$

Step2: Find the first derivative

Use the power rule \((x^{n})^\prime = nx^{n - 1}\).
For \(y = 3x^{2}+9x^{-1}\),
\(\frac{dy}{dx}=(3x^{2})^\prime+(9x^{-1})^\prime\)
\(=3\times2x+9\times(- 1)x^{-2}\)
\(=6x-\frac{9}{x^{2}}\)

Answer:

\(\frac{dy}{dx}=6x-\frac{9}{x^{2}}\)