Question

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

Derivatives
ANSWERED
asked 2020-11-23
Find both first partial derivatives.
\(\displaystyle{f{{\left({x},{y}\right)}}}={x}^{{2}}-{2}{y}^{{2}}+{4}\)

Answers (1)

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:
\(\displaystyle\frac{{\partial{f}}}{{\partial{x}}}={2}{x}\)
\(\displaystyle\frac{{\partial{f}}}{{\partial{y}}}=-{4}{y}\)
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