Question

Find all the second partial derivatives. f(x,y)=x^{4}y-2x^{5}y^{2} f_{xx}(x,y)= f_{xy}(x,y)= f_{yx}(x,y)= f_{yy}(x,y)=

Derivatives
ANSWERED
asked 2021-05-09
Find all the second partial derivatives.
\(\displaystyle{f{{\left({x},{y}\right)}}}={x}^{{{4}}}{y}-{2}{x}^{{{5}}}{y}^{{{2}}}\)
\(\displaystyle{{f}_{{\times}}{\left({x},{y}\right)}}=\)
\(\displaystyle{{f}_{{{x}{y}}}{\left({x},{y}\right)}}=\)
\(\displaystyle{{f}_{{{y}{x}}}{\left({x},{y}\right)}}=\)
\(\displaystyle{{f}_{{{y}{y}}}{\left({x},{y}\right)}}=\)

Answers (1)

2021-05-10
Step 1
Given function is \(\displaystyle{f{{\left({x},{y}\right)}}}={x}^{{{4}}}{y}-{2}{x}^{{{5}}}{y}^{{{2}}}\)
Obtain the first partial derivative with respect to x and y.
\(\displaystyle{{f}_{{{x}}}{\left({x},{y}\right)}}={\left({x}^{{{4}}}{y}-{2}{x}^{{{5}}}{y}^{{{2}}}\right)}'\)
\(\displaystyle={4}{x}^{{{3}}}{y}-{10}{x}^{{{4}}}{y}^{{{2}}}\)
\(\displaystyle{{f}_{{{y}}}{\left({x},{y}\right)}}={\left({x}^{{{4}}}{y}-{2}{x}^{{{5}}}{y}^{{{2}}}\right)}'\)
\(\displaystyle={x}^{{{4}}}-{4}{x}^{{{5}}}{y}\)
Step 2
Compute the second derivative.
\(\displaystyle{{f}_{{\times}}{\left({x},{y}\right)}}={\left({4}{x}^{{{3}}}{y}-{10}{x}^{{{4}}}{y}^{{{2}}}\right)}'\)
\(\displaystyle={12}{x}^{{{2}}}{y}-{40}{x}^{{{3}}}{y}^{{{2}}}\)
\(\displaystyle{{f}_{{{x}{y}}}{\left({x},{y}\right)}}={\left({4}{x}^{{{3}}}{y}-{10}{x}^{{{4}}}{y}^{{{2}}}\right)}'\)
\(\displaystyle={4}{x}^{{{3}}}-{20}{x}^{{{4}}}{y}\)
\(\displaystyle{{f}_{{{y}{x}}}{\left({x},{y}\right)}}={\left({x}^{{{4}}}-{4}{x}^{{{5}}}{y}\right)}'\)
\(\displaystyle={4}{x}^{{{3}}}-{20}{x}^{{{4}}}{y}\)
\(\displaystyle{f}_{{{y}{y}}}={\left({x}^{{{4}}}-{4}{x}^{{{5}}}{y}\right)}'\)
\(\displaystyle={0}-{4}{x}^{{{5}}}\)
\(\displaystyle=-{4}{x}^{{{5}}}\)
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