Alyce Wilkinson

2020-11-23

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

Leonard Stokes

Expert

Step 1
we have to find the first partial derivatives:
the given function is:
$f\left(x,y\right)={x}^{2}-2{y}^{2}+4$
Step 2
$\frac{\partial f}{\partial x}=\frac{\partial \left({x}^{2}-2{y}^{2}+4\right)}{\partial x}$
$=\frac{\partial {x}^{2}}{\partial x}-2\frac{\partial {x}^{2}}{\partial x}+\frac{\partial 4}{\partial x}$
$=2x-2\left(0\right)+0$
$=2x-0+0=2x$
$\frac{\partial f}{\partial y}=\frac{\partial \left({x}^{2}-2{y}^{2}+4\right)}{\partial y}$
$=\frac{\partial {x}^{2}}{\partial y}-\frac{\partial 2{y}^{2}}{\partial y}+\frac{\partial 4}{\partial y}$
$=0-2\frac{\partial {y}^{2}}{\partial y}+0$
$=-2\left(2y\right)=-4y$
Step 3
therefore the value of both first partial derivatives is:
$\frac{\partial f}{\partial x}=2x$
$\frac{\partial f}{\partial y}=-4y$

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