Let \(f(x, y) = tan^{-1} (\frac yx)\)
\(x -1 = 0 \) ⇒ \(x = 1\) ⇒ \(a=1\) (Let)
& \(y -1 = 0 \) ⇒ \(y = 1\) ⇒ \(y=1\) (Let)
Taylor series
\(f(x, y) = f(a, b) + \frac 1{1!} \left[ (x-a) \left(\frac{\partial f}{\partial x}\right)_{(a, b)} +(y-b)\left(\frac{\partial f}{\partial y}\right)_{(a, b)} \right]\)
\(+ \frac 1{2!} \left[(x-a)^2 \left(\frac{\partial^2 f}{\partial x^2}\right)_{(a, b)} + (y-b)^2\left(\frac{\partial^2 f}{\partial y^2}\right)_{(a, b)} + 2(x-a)(y-b) \left(\frac{\partial^2 f}{\partial x\partial y}\right)_{(a, b)}\right]\)
Put \(a = 1, b=1\)
Then
\(f(x, y) = f(1, 1) + \frac 1{1!} \left[ (x-1) \left(\frac{\partial f}{\partial x}\right)_{(1,1)} +(y-1)\left(\frac{\partial f}{\partial y}\right)_{(1, 1)} \right]\)
\(+ \frac 1{2!} \left[(x-1)^2 \left(\frac{\partial^2 f}{\partial x^2}\right)_{(1, 1)} + (y-1)^2\left(\frac{\partial^2 f}{\partial y^2}\right)_{(1, 1)} + 2(x-1)(y-1) \left(\frac{\partial^2 f}{\partial x\partial y}\right)_{(1, 1)}\right]\)
\(+...\)
\(\because f(x, y) = tan^{-1} (\frac yx)\)
\( f(1, 1) = tan^{-1} (\frac 11) = tan^{-1}(1) = \frac \pi4\)
\(\frac{\partial f}{\partial x} = \frac \partial {\partial x} tan^{-1} (\frac yx) = \frac 1{1 + (\frac yx)^2} \frac{\partial }{\partial x}(\frac yx)\)
\(= \frac{x^2}{x^2 + y^2} . \frac{-y}{x^2} =\frac {-y}{x^2 + y^2}\)
\(\frac{\partial f}{\partial y} = \frac \partial {\partial x} tan^{-1} (\frac yx) = \frac 1{1 + (\frac yx)^2} .\frac 1x = \frac x{x^2 + y^2}\)
\(\frac{\partial^2 f}{\partial x^2} = \frac \partial {\partial x} \left(\frac {-y}{x^2 + y^2}\right) = \frac {-y}{(x^2 + y^2)^2} . 2x = \frac{2xy}{(x^2 + y^2)^2}\)
\(\frac{\partial^2 f}{\partial y^2} = \frac \partial {\partial y} \left(\frac {x}{x^2 + y^2}\right) = \frac {-x}{(x^2 + y^2)^2} . 2y = \frac{-2xy}{(x^2 + y^2)^2}\)
\(\frac{\partial^2 f}{\partial x \partial y} = \frac \partial {\partial x} \left(\frac {x}{x^2 + y^2}\right) =\frac{(x^2 + y^2).1-x.2x}{(x^2 + y^2)^2} = \frac{y^2 - x^2}{(x^2 + y^2)^2}\)
\(\therefore \left(\frac{\partial f}{\partial x}\right)_{(1,1)} = \frac{-1}2\)
\(\left(\frac{\partial f}{\partial y}\right)_{(1,1)} = \frac{1}2\)
\( \left(\frac{\partial^2 f}{\partial x^2}\right)_{(1,1)} = \frac24=\frac12\)
\( \left(\frac{\partial^2 f}{\partial y^2}\right)_{(1,1)} = \frac{-2}4=\frac{-1}2\)
\( \left(\frac{\partial^2 f}{\partial x\partial y}\right)_{(1,1)} = 0\)
\(\therefore f(x, y) = \frac \pi4 + \left((x-1) (\frac{-1}2) + (y-1)(\frac 12)\right)\\ + \frac 12 \left((x-1)^2 (\frac 12) + (y-1)^2(\frac{-1}2)+2(x-1)(y-1)\times 0\right)\)
\(= \frac \pi4 +\frac 12 (- x + 1 + y - 1) + \frac 14 ((x-1)^2 - (y-1)^2)\)
\(= \frac \pi4 + \frac{y-x}2 + \frac 14 (x^2 - 2x - y^2 + 2y)\)
