Step 1

we have to find the first partial derivatives:

the given function is:

\(\displaystyle{f{{\left({x},{y}\right)}}}={x}^{{2}}-{2}{y}^{{2}}+{4}\)

Step 2

\(\displaystyle\frac{{\partial{f}}}{{\partial{x}}}=\frac{{\partial{\left({x}^{{2}}-{2}{y}^{{2}}+{4}\right)}}}{{\partial{x}}}\)

\(\displaystyle=\frac{{\partial{x}^{{2}}}}{{\partial{x}}}-{2}\frac{{\partial{x}^{{2}}}}{{\partial{x}}}+\frac{{\partial{4}}}{{\partial{x}}}\)

\(\displaystyle={2}{x}-{2}{\left({0}\right)}+{0}\)

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

\(\displaystyle\frac{{\partial{f}}}{{\partial{y}}}=\frac{{\partial{\left({x}^{{2}}-{2}{y}^{{2}}+{4}\right)}}}{{\partial{y}}}\)

\(\displaystyle=\frac{{\partial{x}^{{2}}}}{{\partial{y}}}-\frac{{\partial{2}{y}^{{2}}}}{{\partial{y}}}+\frac{{\partial{4}}}{{\partial{y}}}\)

\(\displaystyle={0}-{2}\frac{{\partial{y}^{{2}}}}{{\partial{y}}}+{0}\)

\(\displaystyle=-{2}{\left({2}{y}\right)}=-{4}{y}\)

Step 3

therefore the value of both first partial derivatives is:

\(\displaystyle\frac{{\partial{f}}}{{\partial{x}}}={2}{x}\)

\(\displaystyle\frac{{\partial{f}}}{{\partial{y}}}=-{4}{y}\)

we have to find the first partial derivatives:

the given function is:

\(\displaystyle{f{{\left({x},{y}\right)}}}={x}^{{2}}-{2}{y}^{{2}}+{4}\)

Step 2

\(\displaystyle\frac{{\partial{f}}}{{\partial{x}}}=\frac{{\partial{\left({x}^{{2}}-{2}{y}^{{2}}+{4}\right)}}}{{\partial{x}}}\)

\(\displaystyle=\frac{{\partial{x}^{{2}}}}{{\partial{x}}}-{2}\frac{{\partial{x}^{{2}}}}{{\partial{x}}}+\frac{{\partial{4}}}{{\partial{x}}}\)

\(\displaystyle={2}{x}-{2}{\left({0}\right)}+{0}\)

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

\(\displaystyle\frac{{\partial{f}}}{{\partial{y}}}=\frac{{\partial{\left({x}^{{2}}-{2}{y}^{{2}}+{4}\right)}}}{{\partial{y}}}\)

\(\displaystyle=\frac{{\partial{x}^{{2}}}}{{\partial{y}}}-\frac{{\partial{2}{y}^{{2}}}}{{\partial{y}}}+\frac{{\partial{4}}}{{\partial{y}}}\)

\(\displaystyle={0}-{2}\frac{{\partial{y}^{{2}}}}{{\partial{y}}}+{0}\)

\(\displaystyle=-{2}{\left({2}{y}\right)}=-{4}{y}\)

Step 3

therefore the value of both first partial derivatives is:

\(\displaystyle\frac{{\partial{f}}}{{\partial{x}}}={2}{x}\)

\(\displaystyle\frac{{\partial{f}}}{{\partial{y}}}=-{4}{y}\)