Find both first partial derivatives. z = 6x − x^2y + 8y^2

Find both first partial derivatives. z = 6x − x^2y + 8y^2

Question
Derivatives
asked 2021-02-05
Find both first partial derivatives. \(\displaystyle{z}={6}{x}−{x}^{{2}}{y}+{8}{y}^{{2}}\)

Answers (1)

2021-02-06
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,
\(\displaystyle\frac{{\partial{z}}}{{\partial{y}}}=\frac{\partial}{{\partial{y}}}{\left({6}{x}-{x}^{{2}}{y}+{8}{y}^{{2}}\right)}\)
\(\displaystyle=\frac{\partial}{{\partial{y}}}{\left({6}{x}\right)}-\frac{\partial}{{\partial{y}}}{\left({x}^{{2}}{y}\right)}+\frac{\partial}{{\partial{y}}}{\left({8}{y}^{{2}}\right)}\)
\(\displaystyle={0}-{x}^{{2}}+{16}{y}\)
\(\displaystyle={16}{y}-{x}^{{2}}\)
0

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