Now, $ x=e^v sec \; u \; \; \ldots (i) \; \; \; \; \; \; \; \; y=e^v tan \; u \; \; \ldots (ii) \\ \therefore x^2=e^{2v} sec^2 \; u \; \; and x^2=e^{2v} tan^2 \; u \\ x^2 \; - \; y^2 \; = \; e^{2v} \Big( sec^2 \; u - tan^2 \; u \Big) = e^{2v} \{ \because 1 \; + \; tan^2 \; u = sec^2 \; u \; and \; \therefore sec^2 \; u - tan^2 \; u = 1 \} \\ Taking \; log \; on\; both\; sides, \\ \therefore 2v \; = \; log(x^2 - y^2) \; \therefore v \; = \; \frac{1}{2} log(x^2 - y^2) \\ \therefore \frac{\partial v}{\partial x} \; = \; \frac{1}{2} \times \frac{1}{x^2 - y^2} (2x) \; = \; \frac{x}{x^2 - y^2} \\ \; \\ \frac{\partial v}{\partial y} \; = \; \frac{1}{2} \times \frac{1}{x^2 - y^2} (-2y) \; = \; \frac{-y}{x^2 - y^2} \\ \; \\ Equation (ii) divide by (i) \\ \; \\ \therefore \frac{y}{x} \; = \; \frac{e^v tan \; u}{e^v sec \; u} \; = \; \frac{sin \; u / cos \; u}{ 1/ cos\;u} \\ \; \\ u \; = \; sin^{-1} \bigg( \frac{y}{x} \bigg) \; \; \therefore \frac{\partial u}{\partial x} \; = \; \dfrac{1}{\sqrt{1 - \bigg(\frac{y}{x}\bigg)^2}} \frac{\partial }{\partial x}\Big( \frac{y}{x} \Big) \\ \; \\ = \frac{x}{\sqrt{(x^2 \; - \; y^2)}} \Big( \frac{-y}{x^2} \Big) \; = \; \frac{-y}{x\sqrt{(x^2 \; - \; y^2)}} \\ \frac{\partial u}{\partial y} \; = \; \dfrac{1}{\sqrt{1 - \bigg(\frac{y}{x}\bigg)^2}} \frac{\partial }{\partial y}\Big( \frac{y}{x} \Big) \; = \; \frac{x}{\sqrt{(x^2 \; - \; y^2)}} \Big( \frac{1}{x} \Big) \; = \; \frac{1}{\sqrt{(x^2 \; - \; y^2)}} \\ \; \\ J\bigg ( \frac{u,v}{x,y} \bigg) \;= \; \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{-y}{x\sqrt{(x^2 \; - \; y^2)}} & \frac{1}{\sqrt{(x^2 \; - \; y^2)}} \\ \frac{x}{x^2 - y^2} & \frac{-y}{x^2 - y^2} \end{array} \right| \\ = \; \frac{y^2}{x(x^2 - y^2)^{3/2}} \; - \; \frac{x}{(x^2 - y^2)^{3/2}} \\ =\frac{y^2 \; - \; x^2}{x(x^2 - y^2)^{3/2}} \\ = \; \frac{- (x^2 \; - \; y^2)}{x(x^2 - y^2)^{3/2}} \; = \; \frac{- 1 }{x(x^2 - y^2)^{1/2}} \\ = \frac{-1}{xe^v} \; \; \{ \because x^2 \; - \; y^2 \; = \; e^{2v} \} \\ = \frac{-1}{e^v sec\;u \cdot e^v} \\ = \frac{-cos \; u}{e^{2v}} \; = \; -e^{-2v}cos\;u $