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

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

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

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