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

nicekikah 2021-10-03 Answered
Solve the derivatives.
\(\displaystyle{v}={\left({z}^{{{4}}}-{2}{z}+{1}\right)}^{{{\frac{{{3}}}{{{2}}}}}}\)

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Expert Answer

lamusesamuset
Answered 2021-10-04 Author has 17896 answers
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: \(\displaystyle{u}={\left({z}^{{{4}}}-{2}{z}+{1}\right)}\)
\(\displaystyle{\frac{{{d}{v}}}{{{\left.{d}{z}\right.}}}}={\frac{{{3}}}{{{2}}}}{\left({z}^{{{4}}}-{2}{z}+{1}\right)}^{{{\frac{{{1}}}{{{2}}}}}}{\left({4}{z}^{{{3}}}-{2}\right)}\)
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