Using the definition, calculate the derivatives of the functions. Then find the values of the derivatives as specified.

$k\left(z\right)=\frac{1-z}{2z};{k}^{\prime}(-1),{k}^{\prime}\left(1\right),{k}^{\prime}\left(\sqrt{2}\right)$

Brennan Flores
2021-09-28
Answered

Using the definition, calculate the derivatives of the functions. Then find the values of the derivatives as specified.

$k\left(z\right)=\frac{1-z}{2z};{k}^{\prime}(-1),{k}^{\prime}\left(1\right),{k}^{\prime}\left(\sqrt{2}\right)$

You can still ask an expert for help

Laaibah Pitt

Answered 2021-09-29
Author has **98** answers

Step 1

$k\left(z\right)=\frac{1-z}{2z}$

Step 2

Using Quotient Rule

$\left(\frac{f}{g}\right)=\frac{{f}^{\prime}g-{g}^{\prime}f}{{g}^{2}}$

Step 3

Using quotient rule to find the derivative

${k}^{\prime}\left(z\right)=\frac{1}{2}\left[\frac{1-z}{z}\right]$

${k}^{\prime}\left(z\right)=\frac{1}{2}\left[\frac{z{(1-z)}^{\prime}-{z}^{\prime}(1-z)}{{z}^{2}}\right]$

${k}^{\prime}\left(z\right)=\frac{1}{2}\left[\frac{-z-(1-z)}{{z}^{2}}\right]$

${k}^{\prime}\left(z\right)=\frac{1}{2}\left[\frac{-z-1+z}{{z}^{2}}\right]$

${k}^{\prime}\left(z\right)=\left[\frac{1}{2{z}^{2}}\right]$

Step 4

Put z=1

${k}^{\prime}\left(1\right)=\left[\frac{1}{2}\right]$

Put z=-1

${k}^{\prime}(-1)=\left[\frac{1}{2}\right]$

put$z=\sqrt{2}$

${k}^{\prime}\left(\sqrt{2}\right)=\left[\frac{1}{4}\right]$

Step 2

Using Quotient Rule

Step 3

Using quotient rule to find the derivative

Step 4

Put z=1

Put z=-1

put

Jeffrey Jordon

Answered 2021-11-20
Author has **2027** answers

Answer is given below (on video)

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