Step 1: Given that

Find both first partial derivatives. \(\displaystyle{z}={6}{x}−{x}^{{2}}{y}+{8}{y}^{{2}}\)

Step 2: Solve

We have,

\(\displaystyle{z}={6}{x}−{x}^{{2}}{y}+{8}{y}^{{2}}\)

Differentiating partially both sides w.r.t x we obtain,

\(\displaystyle\frac{{\partial{z}}}{{\partial{x}}}=\frac{\partial}{{\partial{x}}}{\left({6}{x}-{x}^{{2}}{y}+{8}{y}^{{2}}\right)}\)

\(\displaystyle=\frac{\partial}{{\partial{x}}}{\left({6}{x}\right)}-\frac{\partial}{{\partial{x}}}{\left({x}^{{2}}{y}\right)}+\frac{\partial}{{\partial{x}}}{\left({8}{y}^{{2}}\right)}\)

\(\displaystyle={6}-{2}{x}{y}+{0}\)

\(\displaystyle={6}-{2}{x}{y}\)

Similarly, Differentiating partially the given function both sides w.r.t y we obtain,

\(\displaystyle\frac{{\partial{z}}}{{\partial{y}}}=\frac{\partial}{{\partial{y}}}{\left({6}{x}-{x}^{{2}}{y}+{8}{y}^{{2}}\right)}\)

\(\displaystyle=\frac{\partial}{{\partial{y}}}{\left({6}{x}\right)}-\frac{\partial}{{\partial{y}}}{\left({x}^{{2}}{y}\right)}+\frac{\partial}{{\partial{y}}}{\left({8}{y}^{{2}}\right)}\)

\(\displaystyle={0}-{x}^{{2}}+{16}{y}\)

\(\displaystyle={16}{y}-{x}^{{2}}\)

Find both first partial derivatives. \(\displaystyle{z}={6}{x}−{x}^{{2}}{y}+{8}{y}^{{2}}\)

Step 2: Solve

We have,

\(\displaystyle{z}={6}{x}−{x}^{{2}}{y}+{8}{y}^{{2}}\)

Differentiating partially both sides w.r.t x we obtain,

\(\displaystyle\frac{{\partial{z}}}{{\partial{x}}}=\frac{\partial}{{\partial{x}}}{\left({6}{x}-{x}^{{2}}{y}+{8}{y}^{{2}}\right)}\)

\(\displaystyle=\frac{\partial}{{\partial{x}}}{\left({6}{x}\right)}-\frac{\partial}{{\partial{x}}}{\left({x}^{{2}}{y}\right)}+\frac{\partial}{{\partial{x}}}{\left({8}{y}^{{2}}\right)}\)

\(\displaystyle={6}-{2}{x}{y}+{0}\)

\(\displaystyle={6}-{2}{x}{y}\)

Similarly, Differentiating partially the given function both sides w.r.t y we obtain,

\(\displaystyle\frac{{\partial{z}}}{{\partial{y}}}=\frac{\partial}{{\partial{y}}}{\left({6}{x}-{x}^{{2}}{y}+{8}{y}^{{2}}\right)}\)

\(\displaystyle=\frac{\partial}{{\partial{y}}}{\left({6}{x}\right)}-\frac{\partial}{{\partial{y}}}{\left({x}^{{2}}{y}\right)}+\frac{\partial}{{\partial{y}}}{\left({8}{y}^{{2}}\right)}\)

\(\displaystyle={0}-{x}^{{2}}+{16}{y}\)

\(\displaystyle={16}{y}-{x}^{{2}}\)