Question

# Find both first partial derivatives. f(x,y)=x^2-2y^2+4

Derivatives
Find both first partial derivatives.
$$\displaystyle{f{{\left({x},{y}\right)}}}={x}^{{2}}-{2}{y}^{{2}}+{4}$$

2020-11-24
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:
$$\displaystyle\frac{{\partial{f}}}{{\partial{x}}}={2}{x}$$
$$\displaystyle\frac{{\partial{f}}}{{\partial{y}}}=-{4}{y}$$