Question

# Find the partial derivatives, fx and fy, for f(x,y)=8x^{2}-6xy^{2}+3y+21

Derivatives
Find the partial derivatives, fx and fy, for $$\displaystyle{f{{\left({x},{y}\right)}}}={8}{x}^{{{2}}}-{6}{x}{y}^{{{2}}}+{3}{y}+{21}$$

2021-05-05
Step 1
Given:
$$\displaystyle{f{{\left({x},{y}\right)}}}={8}{x}^{{{2}}}-{6}{x}{y}^{{{2}}}+{3}{y}+{21}$$
Step 2
To find:
$$\displaystyle{\frac{{\partial{f{{\left({x},{y}\right)}}}}}{{\partial{x}}}},{\frac{{\partial{f{{\left({x},{y}\right)}}}}}{{\partial{y}}}}$$
Step 3
Calculation:
$$\displaystyle{f{{\left({x},{y}\right)}}}={8}{x}^{{{2}}}-{6}{x}{y}^{{{2}}}+{3}{y}+{21}$$
$$\displaystyle{\frac{{\partial{f{{\left({x},{y}\right)}}}}}{{\partial{x}}}}={\frac{{\partial}}{{\partial{x}}}}{\left({8}{x}^{{{2}}}-{6}{x}{y}^{{{2}}}+{3}{y}+{21}\right)}$$
$$\displaystyle={16}{x}-{6}{y}^{{{2}}}$$
$$\displaystyle{\frac{{\partial{f{{\left({x},{y}\right)}}}}}{{\partial{y}}}}={\frac{{\partial}}{{\partial{y}}}}{\left({8}{x}^{{{2}}}-{6}{x}{y}^{{{2}}}+{3}{y}+{21}\right)}$$
$$\displaystyle=-{12}{x}{y}+{3}$$ Step 4 Answer: $$\displaystyle{\frac{{\partial{f{{\left({x},{y}\right)}}}}}{{\partial{x}}}}={16}{x}-{6}{y}^{{{2}}}$$
$$\displaystyle{\frac{{\partial{f{{\left({x},{y}\right)}}}}}{{\partial{y}}}}=-{12}{x}{y}+{3}$$