Question

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

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

2021-06-05
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 $$\displaystyle{z}=\sqrt{{{2}}}$$
$$\displaystyle{k}'{\left(\sqrt{{{2}}}\right)}={\left[{\frac{{{1}}}{{{4}}}}\right]}$$