# Using the definition, calculate the derivatives of the functions. Then

Using the definition, calculate the derivatives of the functions. Then find the values of the derivatives as specified.
$k\left(z\right)=\frac{1-z}{2z};{k}^{\prime }\left(-1\right),{k}^{\prime }\left(1\right),{k}^{\prime }\left(\sqrt{2}\right)$
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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
${k}^{\prime }\left(z\right)=\frac{1}{2}\left[\frac{1-z}{z}\right]$
${k}^{\prime }\left(z\right)=\frac{1}{2}\left[\frac{z{\left(1-z\right)}^{\prime }-{z}^{\prime }\left(1-z\right)}{{z}^{2}}\right]$
${k}^{\prime }\left(z\right)=\frac{1}{2}\left[\frac{-z-\left(1-z\right)}{{z}^{2}}\right]$
${k}^{\prime }\left(z\right)=\frac{1}{2}\left[\frac{-z-1+z}{{z}^{2}}\right]$
${k}^{\prime }\left(z\right)=\left[\frac{1}{2{z}^{2}}\right]$
Step 4
Put z=1
${k}^{\prime }\left(1\right)=\left[\frac{1}{2}\right]$
Put z=-1
${k}^{\prime }\left(-1\right)=\left[\frac{1}{2}\right]$
put $z=\sqrt{2}$
${k}^{\prime }\left(\sqrt{2}\right)=\left[\frac{1}{4}\right]$
Jeffrey Jordon