Question

find the first and second derivatives of the function.

$w = \left( \frac { 4 + 9 z } { 3 z } \
ight) ( 6 - z )$

$\frac { d w } { d z } = - \frac { 8 } { z ^ { 2 } } - 3$

$\frac { d ^ { 2 } w } { d z ^ { 2 } } = \square$

Explanation:

Step1: Differentiate the first - derivative

We have \(\frac{dw}{dz}=-\frac{8}{z^{2}}-3=-8z^{- 2}-3\)
Using the power rule \(\frac{d}{dz}(z^{n})=nz^{n - 1}\)
For the term \(-8z^{-2}\), when we differentiate with respect to \(z\):
\(\frac{d}{dz}(-8z^{-2})=-8\times(-2)z^{-2 - 1}\)
For the term \(-3\) (a constant), \(\frac{d}{dz}(-3) = 0\)

Step2: Simplify the result

\(\frac{d}{dz}(-8z^{-2})=16z^{-3}=\frac{16}{z^{3}}\) and \(\frac{d}{dz}(-3) = 0\)

Answer:

\(\frac{d^{2}w}{dz^{2}}=\frac{16}{z^{3}}\)