# Solve the derivatives. v=(z^{4}-2z+1)^{\frac{3}{2}}

Solve the derivatives.
$$\displaystyle{v}={\left({z}^{{{4}}}-{2}{z}+{1}\right)}^{{{\frac{{{3}}}{{{2}}}}}}$$

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Step 1
Given:
$$\displaystyle{v}={\left({z}^{{{4}}}-{2}{z}+{1}\right)}^{{{\frac{{{3}}}{{{2}}}}}}$$
Step 2
$$\displaystyle{v}={\left({z}^{{{4}}}-{2}{z}+{1}\right)}^{{{\frac{{{3}}}{{{2}}}}}}$$
Differentiate with respect to z
$$\displaystyle{\frac{{{d}{v}}}{{{\left.{d}{z}\right.}}}}={\frac{{{d}}}{{{\left.{d}{z}\right.}}}}{\left[{\left({z}^{{{4}}}-{2}{z}+{1}\right)}^{{{\frac{{{3}}}{{{2}}}}}}\right]}$$
Apply chain rule: $$\displaystyle{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left({f{{\left({g{{\left({x}\right)}}}\right)}}}\right)}={\frac{{{d}}}{{{d}{u}}}}{f{{\left({u}\right)}}}{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{g{{\left({x}\right)}}}\ {w}{i}{t}{h}\ {u}={\left({z}^{{{4}}}-{2}{z}+{1}\right)}$$
$$\displaystyle{\frac{{{d}{v}}}{{{\left.{d}{z}\right.}}}}={\frac{{{d}}}{{{d}{u}}}}{u}^{{{\frac{{{3}}}{{{2}}}}}}{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left({z}^{{{4}}}-{2}{z}+{1}\right)}$$
Apply power rule: $$\displaystyle{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left({x}^{{{n}}}\right)}={n}{x}^{{{n}-{1}}}$$
$$\displaystyle{\frac{{{d}{v}}}{{{\left.{d}{z}\right.}}}}={\frac{{{3}}}{{{2}}}}{u}^{{{\frac{{{3}}}{{{2}}}}-{1}}}{\left({4}{z}^{{{4}-{1}}}-{2}+{0}\right)}$$
$$\displaystyle{\frac{{{d}{v}}}{{{\left.{d}{z}\right.}}}}={\frac{{{3}}}{{{2}}}}{u}^{{{\frac{{{1}}}{{{2}}}}}}{\left({4}{z}^{{{3}}}-{2}\right)}$$
Put back: $$\displaystyle{u}={\left({z}^{{{4}}}-{2}{z}+{1}\right)}$$
$$\displaystyle{\frac{{{d}{v}}}{{{\left.{d}{z}\right.}}}}={\frac{{{3}}}{{{2}}}}{\left({z}^{{{4}}}-{2}{z}+{1}\right)}^{{{\frac{{{1}}}{{{2}}}}}}{\left({4}{z}^{{{3}}}-{2}\right)}$$