Step 1

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

Step 2

Using Quotient Rule

\(\displaystyle{\left({\frac{{{f}}}{{{g}}}}\right)}={\frac{{{f}'{g}-{g}'{f}}}{{{g}^{{{2}}}}}}\)

Step 3

Using quotient rule to find the derivative

\(\displaystyle{k}'{\left({z}\right)}={\frac{{{1}}}{{{2}}}}{\left[{\frac{{{1}-{z}}}{{{z}}}}\right]}\)

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

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

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

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

Step 4

Put z=1

\(\displaystyle{k}'{\left({1}\right)}={\left[{\frac{{{1}}}{{{2}}}}\right]}\)

Put z=-1

\(\displaystyle{k}'{\left(-{1}\right)}={\left[{\frac{{{1}}}{{{2}}}}\right]}\)

put \(\displaystyle{z}=\sqrt{{{2}}}\)

\(\displaystyle{k}'{\left(\sqrt{{{2}}}\right)}={\left[{\frac{{{1}}}{{{4}}}}\right]}\)

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

Step 2

Using Quotient Rule

\(\displaystyle{\left({\frac{{{f}}}{{{g}}}}\right)}={\frac{{{f}'{g}-{g}'{f}}}{{{g}^{{{2}}}}}}\)

Step 3

Using quotient rule to find the derivative

\(\displaystyle{k}'{\left({z}\right)}={\frac{{{1}}}{{{2}}}}{\left[{\frac{{{1}-{z}}}{{{z}}}}\right]}\)

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

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

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

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

Step 4

Put z=1

\(\displaystyle{k}'{\left({1}\right)}={\left[{\frac{{{1}}}{{{2}}}}\right]}\)

Put z=-1

\(\displaystyle{k}'{\left(-{1}\right)}={\left[{\frac{{{1}}}{{{2}}}}\right]}\)

put \(\displaystyle{z}=\sqrt{{{2}}}\)

\(\displaystyle{k}'{\left(\sqrt{{{2}}}\right)}={\left[{\frac{{{1}}}{{{4}}}}\right]}\)