Question

Using the definition, calculate the derivatives of the functions. Then find the values of the derivatives as specified.
\(\displaystyle{k}{\left({z}\right)}={\frac{{{1}-{z}}}{{{2}{z}}}};{k}'{\left(-{1}\right)},{k}'{\left({1}\right)},{k}'{\left(\sqrt{{{2}}}\right)}\)

Answer 1

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

Answer 2

Answer is given below (on video)