Answer:
Step-by-step explanation:
[tex]f(x,y)=arctan(xy)\\\\\dfrac{ \partial f}{ \partial x}=\dfrac{y}{1+x^2y^2} \\\\\dfrac{ \partial f}{ \partial y}=\dfrac{x}{1+x^2y^2} \\\\\dfrac{ \partial ^2 f}{ \partial x^2}=\dfrac{-2xy^3}{(1+x^2y^2)^2} \\\\\dfrac{ \partial ^2 f}{ \partial y^2}=\dfrac{-2x^3y}{(1+x^2y^2)^2} \\\\\dfrac{ \partial ^2 f}{ \partial x\ \partial y}=\dfrac{1-x^2y^2}{(1+x^2y^2)^2} \\[/tex]
[tex]f(x,y)= f(1,1)+(x-1)\dfrac{\partial f}{\partial x} (1,1)+(y-1)\dfrac{\partial f}{\partial y} (1,1)+\dfrac{(x-1)^2}{2} \dfrac{\partial^2 f}{\partial x^2} (1,1)+(x-1)(y-1)\dfrac{\partial^2 f}{\partial x\ \partial y} (1,1)+\dfrac{(y-1)^2}{2} \dfrac{\partial^2 f}{\partial y^2} (1,1)+...\\\\=\dfrac{\pi}{4}+\dfrac{(x-1)}{2} +\dfrac{(y-1)}{2} -\dfrac{(x-1)^2}{4} +(x-1)(y-1)*\dfrac{0}{4}-\dfrac{(y-1)^2}{4} \\\\[/tex]
[tex]\boxed{f(x,y)=\dfrac{\pi}{4}+\dfrac{(x-1)}{2} +\dfrac{(y-1)}{2} -\dfrac{(x-1)^2}{4} -\dfrac{(y-1)^2}{4} }\\\\[/tex]
