Alyce Wilkinson

2020-11-23

Find both first partial derivatives.

$f(x,y)={x}^{2}-2{y}^{2}+4$

Leonard Stokes

Skilled2020-11-24Added 98 answers

Step 1

we have to find the first partial derivatives:

the given function is:

$f(x,y)={x}^{2}-2{y}^{2}+4$

Step 2

$\frac{\partial f}{\partial x}=\frac{\partial ({x}^{2}-2{y}^{2}+4)}{\partial x}$

$=\frac{\partial {x}^{2}}{\partial x}-2\frac{\partial {x}^{2}}{\partial x}+\frac{\partial 4}{\partial x}$

$=2x-2\left(0\right)+0$

$=2x-0+0=2x$

$\frac{\partial f}{\partial y}=\frac{\partial ({x}^{2}-2{y}^{2}+4)}{\partial y}$

$=\frac{\partial {x}^{2}}{\partial y}-\frac{\partial 2{y}^{2}}{\partial y}+\frac{\partial 4}{\partial y}$

$=0-2\frac{\partial {y}^{2}}{\partial y}+0$

$=-2\left(2y\right)=-4y$

Step 3

therefore the value of both first partial derivatives is:

$\frac{\partial f}{\partial x}=2x$

$\frac{\partial f}{\partial y}=-4y$

we have to find the first partial derivatives:

the given function is:

Step 2

Step 3

therefore the value of both first partial derivatives is: