Question

3.3c differentiation rules highe
(6 points)
calculate the second and the third derivative of $y = 7x - \frac{4}{x}$
$y =$
$y =$
note: you can earn partial credit on this problem.
note: you are in the reduced scoring period. all work counts for 85% of the
preview my answers submit answers
you have attempted this problem 0 times.
you have 5 attempts remaining.

Explanation:

Step1: Find the first derivative

Use the power rule \((x^n)^\prime = nx^{n - 1}\). Rewrite \(y = 7x-\frac{4}{x}=7x - 4x^{-1}\).
The first derivative \(y^\prime=(7x - 4x^{-1})^\prime\).
By the sum - rule \((u + v)^\prime=u^\prime + v^\prime\) and power rule: \(y^\prime=7+4x^{-2}\).

Step2: Find the second derivative

Differentiate \(y^\prime = 7 + 4x^{-2}\) with respect to \(x\).
Using the power rule \((x^n)^\prime=nx^{n - 1}\), \((7)^\prime = 0\) and \((4x^{-2})^\prime=4\times(-2)x^{-3}\).
So \(y^{\prime\prime}=-8x^{-3}=-\frac{8}{x^{3}}\).

Step3: Find the third derivative

Differentiate \(y^{\prime\prime}=-8x^{-3}\) with respect to \(x\).
Using the power rule \((x^n)^\prime=nx^{n - 1}\), \((-8x^{-3})^\prime=-8\times(-3)x^{-4}\).
So \(y^{\prime\prime\prime}=24x^{-4}=\frac{24}{x^{4}}\).

Answer:

\(y^{\prime\prime}=-\frac{8}{x^{3}}\), \(y^{\prime\prime\prime}=\frac{24}{x^{4}}\)