and 4 others joined a min ago.
0
24kviews
Jacobians : If $ u = \frac{x+y}{1-xy} , \; v \; = \; tan^{-1}x \; + \; tan^{-1}y , \; Find \frac{\partial (u,v)}{\partial (x,y)} $
1 Answer
6
5.5kviews
written 7.9 years ago by
teamques10
★ 65k
|
• modified 7.9 years ago |
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} \; = …
and 5 others joined a min ago.
ADD COMMENT
EDIT
Please log in to add an answer.