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

Find both first partial derivatives. $z=6x-{x}^{2}y+8{y}^{2}$
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Elberte
Step 1: Given that
Find both first partial derivatives. $z=6x-{x}^{2}y+8{y}^{2}$
Step 2: Solve
We have,
$z=6x-{x}^{2}y+8{y}^{2}$
Differentiating partially both sides w.r.t x we obtain,
$\frac{\partial z}{\partial x}=\frac{\partial }{\partial x}\left(6x-{x}^{2}y+8{y}^{2}\right)$
$=\frac{\partial }{\partial x}\left(6x\right)-\frac{\partial }{\partial x}\left({x}^{2}y\right)+\frac{\partial }{\partial x}\left(8{y}^{2}\right)$
$=6-2xy+0$
$=6-2xy$
Similarly, Differentiating partially the given function both sides w.r.t y we obtain,
$\frac{\partial z}{\partial y}=\frac{\partial }{\partial y}\left(6x-{x}^{2}y+8{y}^{2}\right)$
$=\frac{\partial }{\partial y}\left(6x\right)-\frac{\partial }{\partial y}\left({x}^{2}y\right)+\frac{\partial }{\partial y}\left(8{y}^{2}\right)$
$=0-{x}^{2}+16y$
$=16y-{x}^{2}$