# How do you find the instantaneous rate of change of w with respect to z for w=1/z+z/2?

Sep 28, 2014

$\frac{\mathrm{dw}}{\mathrm{dz}} = - \frac{1}{z} ^ 2 + \frac{1}{2}$

Explanation :

$\frac{\mathrm{dw}}{\mathrm{dz}} = \frac{d}{\mathrm{dz}} \left(\frac{1}{z} + \frac{z}{2}\right)$

Initial set-up.

$\frac{\mathrm{dw}}{\mathrm{dz}} = \frac{d}{\mathrm{dz}} \left(\frac{1}{z}\right) + \frac{d}{\mathrm{dz}} \left(\frac{z}{2}\right)$

The derivative of a sum is equal to the sum of the derivatives.

$\frac{\mathrm{dw}}{\mathrm{dz}} = \frac{d}{\mathrm{dz}} \left({z}^{-} 1\right) + \frac{1}{2} \frac{d}{\mathrm{dz}} \left(z\right)$

First part: A function $f \left(z\right) = \frac{c}{{z}^{n}}$ with $c$ constant can also be written as $f \left(z\right) = c {z}^{- n}$ Second part: $\frac{d}{\mathrm{dz}} c f \left(z\right) = c \frac{d}{\mathrm{dz}} f \left(z\right)$ if c is constant.

$\frac{\mathrm{dw}}{\mathrm{dz}} = - 1 \cdot {z}^{-} 2 + \frac{1}{2} \cdot 1$

Use of the power rule: $\frac{d}{\mathrm{dz}} {z}^{n} = n {z}^{n - 1}$. Then $\frac{d}{\mathrm{dz}} z = \frac{d}{\mathrm{dz}} {z}^{1} = {z}^{0} = 1$

$\frac{\mathrm{dw}}{\mathrm{dz}} = - {z}^{-} 2 + \frac{1}{2}$

Multiplicative identity postulate.

$\frac{\mathrm{dw}}{\mathrm{dz}} = - \frac{1}{z} ^ 2 + \frac{1}{2}$

A function written as $f \left(z\right) = c {z}^{- n}$ can also be written $f \left(z\right) = \frac{c}{{z}^{n}}$