If x = uv, y `=(u+v)/(u-v).`find `(del(u,v))/(del(x,y))`.
Solution
`(del(u,v))/(del(x,y))= `\[\begin{vmatrix}u_x & u_y \\ v_x & v_y\end{vmatrix}\] =\[ \begin{vmatrix} \frac{\delta x}{\delta u} & \frac{\delta x}{\delta v} \\ \frac{\delta y}{\delta u} & \frac{\delta y}{\delta v}\end{vmatrix}\]
`(del x)/(del u)=del(uv)=v.`..................(2)
`(delx)/(delv)=del(uv)=u.`....................(3)
`(dely)/(delu)=del((u+v)/(u-v))=((u-v)-(u+v))/((u-v)^2)=(-2v)/(u-v)^2`...............(4)
`(dely)/(delv)=del((u+v)/(u-v))=((u-v)+(u+v))/(u-v)^2=(2u)/((u-v)^2` ....................(5)
From equation (2),(3),(4),(5) we get,
\[\begin{vmatrix} \frac{\delta x}{\delta u} & \frac{\delta x}{\delta v} \\ \frac{\delta y}{\delta u} & \frac{\delta y}{\delta v}\end{vmatrix}\] = \[\begin{vmatrix} v & u \\ \frac {-2v} {u-v}^2 & \frac {2u} {uv} \end{vmatrix}\] = `(2uv)/(u-v)^2+(2uv)/(u-v)^2=(4uv)/(u-v)^2`
From (1) we get,
JJ’=1
J`xx(4uv)/(u-v)^2=1` .........................(let J'=`(4uv)/(u-v)^2`)
Hence J = `(u-v)^2/(4uv)`
`therefore e^(2varphi)=cot (alpha/2)`
