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)`