Question

Find all first and the second partial derivatives. f(x,y)=2x^5y^2+x^2y

Derivatives
ANSWERED
asked 2021-03-02
Find all first and the second partial derivatives.
\(\displaystyle{f{{\left({x},{y}\right)}}}={2}{x}^{{5}}{y}^{{2}}+{x}^{{2}}{y}\)

Answers (1)

2021-03-03
Step 1 Let us find first and second order partial derivatives.
First order partial derivatives:
\(\displaystyle{f}_{{x}}=\frac{{\partial{f}}}{{\partial{x}}}{f}_{{y}}=\frac{{\partial{f}}}{{\partial{y}}}\)
Second order partial derivatives:
\(\displaystyle{f}_{{\times}}=\frac{{\partial^{{2}}{f}}}{{\partial{x}^{{2}}}}=\frac{\partial}{{\partial{x}}}{\left(\frac{\partial}{{\partial{x}}}\right)},{f}_{{{y}{y}}}=\frac{{\partial^{{2}}{f}}}{{\partial{y}^{{2}}}}=\frac{\partial}{{\partial{y}}}{\left(\frac{\partial}{{\partial{y}}}\right)}\)
Step 2 Answer will be at end of step 2
\(\displaystyle{f{{\left({x},{y}\right)}}}={2}{x}^{{5}}{y}^{{2}}+{x}^{{2}}{y}\)
First ordrer partial derivative:
\(\displaystyle{f}_{{x}}=\frac{{\partial{f}}}{{\partial{x}}}={2}.{\left({5}\right)}{x}^{{{5}-{1}}}{y}^{{2}}+{2}{x}^{{{2}-{1}}}{y}\)
\(\displaystyle{f}_{{x}}={10}{x}^{{4}}{y}^{{2}}+{2}{x}{y}\)
\(\displaystyle{f}_{{y}}=\frac{{\partial{f}}}{{\partial{y}}}={2}{x}^{{5}}.{\left({2}\right)}{y}^{{{2}-{1}}}+{x}^{{2}}\)
\(\displaystyle{f}_{{y}}={4}{x}^{{5}}{y}+{x}^{{2}}\)
Second order partial derivative:
\(\displaystyle{f}_{{\times}}\frac{\partial}{{\partial{x}}}{\left({10}{x}^{{4}}{y}^{{2}}+{2}{x}{y}\right)}\)
\(\displaystyle{f}_{{\times}}{4}.{\left({10}\right)}{x}^{{3}}{y}^{{2}}+{2}{y}\)
\(\displaystyle{f}_{{\times}}={40}{x}^{{3}}{y}^{{2}}+{2}{y}\)
\(\displaystyle{f}_{{{y}{y}}}=\frac{\partial}{{\partial{y}}}{\left({4}{x}^{{5}}{y}+{x}^{{2}}\right)}\)
\(\displaystyle{f}_{{{y}{y}}}={4}{x}^{{5}}\)
Step 3
Result:
\(\displaystyle{f}_{{x}}={10}{x}^{{4}}{y}^{{2}}+{2}{x}{y}\)
\(\displaystyle{f}_{{y}}={4}{x}^{{5}}{y}+{x}^{{2}}\)
\(\displaystyle{f}_{{\times}}={40}{x}^{{3}}{y}^{{2}}+{2}{y}\)
\(\displaystyle{f}_{{{y}{y}}}={4}{x}^{{5}}\)
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