Find the first partial derivatives of f(x,y)=sin(x−y) at the point (-9, -9).

[tex]\dfrac{\partial f}{\partial x}=\cos(x-y)[/tex]

[tex]\dfrac{\partial f}{\partial y}=-\cos(x-y)[/tex]

You have [tex]\cos(-9-(-9))=\cos0=1[/tex], so the partial derivative with respect to [tex]x[/tex] is [tex]1[/tex], while the partial derivative with respect to [tex]y[/tex] is [tex]-1[/tex].

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ibiscollingwood ibiscollingwood · Answer 2 · 2022-03-04T12:19:03+01:00

The first partial derivatives of given function with respect to x is [tex]1[/tex] and with respect to y is [tex]-1[/tex]

Partial derivative:

The given function is,

[tex]f(x,y)=sin(x-y)[/tex]

first partial derivatives of given function with respect to x is,

[tex]\frac{\partial f}{\partial x}=cos(x-y)[/tex]

At point (-9, -9)

[tex]\frac{\partial f}{\partial x}=cos(-9+9)=cos(0)=1[/tex]

first partial derivatives of given function with respect to y is,

[tex]\frac{\partial f}{\partial x}=-cos(x-y)[/tex]

At point (-9, -9),

[tex]\frac{\partial f}{\partial x}=-cos(-9+9)=-cos(0)=-1[/tex]

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