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Jacobians : If $ u = \frac{x+y}{1-xy} , \; v \; = \; tan^{-1}x \; + \; tan^{-1}y , \; Find \frac{\partial (u,v)}{\partial (x,y)} $
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u$ = \dfrac{x+y}{1-xy} \; \; \therefore \dfrac{\partial u }{\partial x} \; = \; \frac{(1-xy) \dfrac{\partial }{\partial x}(x+y) \; - \; (x+y)\dfrac{\partial }{\partial x}(1-xy)}{ (1-xy)^2} \; \; \; \; \ldots diff. \; partially \; w.r.t \; x \\ \; \\ \therefore \dfrac{\partial u }{\partial x} \; = \; \frac{(1-xy)(1)-(x+y)(-y)}{(1-xy)^2} \; = \; \frac{1 - xy + y^2 + xy}{(1-xy)^2} \; = \; \frac{ 1 + y^2}{(1-xy)^2} \\ \; \\ \; \\ \dfrac{\partial u }{\partial y} \; = \; \frac{(1-xy) \dfrac{\partial }{\partial y}(x+y) \; - \; (x+y)\dfrac{\partial }{\partial y}(1-xy)}{ (1-xy)^2} \; = \; \frac{(1-xy)(1)-(x+y)(-x)}{(1-xy)^2} \\ \; \\ = \frac{1 - xy + x^2 + xy}{(1-xy)^2} \\ \; \\ \therefore \dfrac{\partial u }{\partial y} \; = \; \frac{ 1 + x^2}{(1-xy)^2} \\ \; \\ \; \\ v \; = \; tan^{-1}x \; + \; tan^{-1}y \\ \; \\ \therefore \frac{\partial v }{\partial x} \; = \; \frac{1}{1+x^2} \\ \; \\ \therefore \frac{\partial v }{\partial y} \; = \; \frac{1}{1+y^2} \\ \; \\ \; \\ Jocobian|J| \; = \; \frac{\partial (u,v)}{\partial (x,y)} \;= \; \left| \begin{array}{ccc} \frac{\partial u }{\partial x} & \frac{\partial u }{\partial y} \\ \frac{\partial v }{\partial x} & \frac{\partial v }{\partial y} \end{array} \right| \;= \; \left| \begin{array}{ccc} \frac{ 1 + y^2}{(1-xy)^2} & \frac{ 1 + x^2}{(1-xy)^2} \\ \frac{1}{1+x^2} & \frac{1}{1+y^2} \end{array} \right| \\ \; \\ \frac{\partial (u,v)}{\partial (u,v)} \;= \; \frac{ 1 + y^2}{(1-xy)^2} \cdot \frac{1}{1+y^2} \; - \; \frac{ 1 + x^2}{(1-xy)^2} \cdot \frac{1}{1+x^2} \\ \; \\ = \frac{ 1 }{(1-xy)^2} \; - \; \frac{ 1 + y^2}{(1-xy)^2} \; = \; 0 \\ \; \\ \; \\ \frac{\partial (u,v)}{\partial (x,y)} \;= \; 0 $

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