Compute the derivatives indicated. f(x,y)=3x^{2}y-6xy^{4}, \frac{\partial

beljuA 2021-10-01 Answered
Compute the derivatives indicated.
\(\displaystyle{f{{\left({x},{y}\right)}}}={3}{x}^{{{2}}}{y}-{6}{x}{y}^{{{4}}},{\frac{{\partial^{{{2}}}{f}}}{{\partial{x}^{{{2}}}}}}\ {\quad\text{and}\quad}\ {\frac{{\partial^{{{2}}}{f}}}{{\partial{y}^{{{2}}}}}}\)

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Expert Answer

Sadie Eaton
Answered 2021-10-02 Author has 18652 answers

Step 1
Here given as
\(\displaystyle{f{{\left({x},{y}\right)}}}={3}{x}^{{{2}}}{y}-{6}{x}{y}^{{{4}}}\)
Now
\(\displaystyle{\frac{{\partial{f}}}{{\partial{x}}}}={\frac{{\partial}}{{\partial{x}}}}{\left({3}{x}^{{{2}}}{y}-{6}{x}{y}^{{{4}}}\right)}\)
\(\displaystyle={\left({6}{x}{y}-{6}{y}^{{{4}}}\right)}\)
\(\displaystyle{\frac{{\partial^{{{2}}}{f}}}{{\partial{x}^{{{2}}}}}}={\frac{{\partial}}{{\partial{x}}}}{\left({6}{x}{y}-{6}{y}^{{{4}}}\right)}\)
\(\displaystyle\Rightarrow{\frac{{\partial^{{{2}}}{f}}}{{\partial{x}^{{{2}}}}}}={6}{y}\)
Step 2
Thus
\(\displaystyle{\frac{{\partial{f}}}{{\partial{y}}}}={\frac{{\partial}}{{\partial{y}}}}{\left({3}{x}^{{{2}}}{y}-{6}{x}{y}^{{{4}}}\right)}\)
\(\displaystyle={\left({3}{x}^{{{2}}}-{24}{x}{y}^{{{3}}}\right)}\)
\(\displaystyle{\frac{{\partial^{{{2}}}{f}}}{{\partial{y}^{{{2}}}}}}={\frac{{\partial}}{{\partial{y}}}}{\left({3}{x}^{{{2}}}-{24}{x}{y}^{{{3}}}\right)}\)
\(\displaystyle\Rightarrow{\frac{{\partial^{{{2}}}{f}}}{{\partial{y}^{{{2}}}}}}-{62}{x}{y}^{{{2}}}\)

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