Verified Answer :

LammettHash

\(\dfrac{\partial f}{\partial x}=\cos(x-y)\)

\(\dfrac{\partial f}{\partial y}=-\cos(x-y)\)

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

The first partial derivatives of given function with respect to x is \(1\) and with respect to y is \(-1\)

Partial derivative:

The given function is,

                    \(f(x,y)=sin(x-y)\)

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

                   \(\frac{\partial f}{\partial x}=cos(x-y)\)

At point (-9, -9)

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

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

                  \(\frac{\partial f}{\partial x}=-cos(x-y)\)

At point (-9, -9),

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

Learn more about the derivative of function here:

https://brainly.com/question/12047216

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