Verifing the solution

Question

$\mathbf{Question:}$ Let $f(x,y)=e^{\sin(x-y)}$ for $(x,y)\in \Bbb R^2$

Find the affine function that is a first order approximation to the function $f$ at the point $(0,0)$

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$\mathbf{Answer:}$

Let's define a function $g(x,y):\Bbb R^2 \to \Bbb R$

I Will use taylor theorem, and I Will get

$g(x,y)= 1+ 1.e^{0}((x,y)-(0,0))=1+(x,y)$ is first order approximation of $f$

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Is this answer right and enough? Please give me a feedback. Thank you for helping:)

Answer 1

I get $$\partial_x g|_{(x,y)=(0,0)} = e^{\sin (x-y)}\cos (x-y)|_{(x,y)=(0,0)}=1$$ $$\partial_y g|_{(x,y)=(0,0)} = e^{\sin (x-y)}\cos (x-y)(-1)|_{(x,y)=(0,0)}=-1,$$so the affine function is $1 + x - y$.