Answer to Question #261656 in Calculus for haemha

Question

 If z = f(x, y), where x = e^u + e^−v , y = e^−u − e^v , then find ∂(u,v)/∂(x,y) .

Answer 1

Consider: $z=f(x,y),$ where $x=e^u+e^{-v},y=e^{-u}-e^v$

Require to find: $\frac{\partial (u,v)}{\partial (x,y)}$

Recollect the following: $\frac{\partial (u,v)}{\partial (x,y)}=\frac{1}{\frac{\partial (x,y)}{\partial (u,v)}}$ and $\frac{\partial (x,y)}{\partial (u,v)}=\begin{vmatrix} \frac{\partial x}{\partial u}&\frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u}& \frac{\partial y}{\partial v} \end{vmatrix}$

Now $x=e^u+e^{-v},y=e^{-u}-e^v\Rightarrow \frac{\partial (x,y)}{\partial (u,v)}=\begin{vmatrix} e^u & -e^{-v}\\ -e^{-u}& -e^v \end{vmatrix}$

$\Rightarrow \frac{\partial (x,y)}{\partial (u,v)}=-e^{u+v}-e^{-u-v}$

Using the above, we have

$\frac{\partial (u,v)}{\partial (x,y)}=\frac{1}{\frac{\partial (x,y)}{\partial (u,v)}}\Rightarrow \frac{\partial (u,v)}{\partial (x,y)}=\frac{1}{-e^{u+v}-e^{-u-v}}$

Therefore,

$\frac{\partial (u,v)}{\partial (x,y)}=\frac{1}{-e^{u+v}-e^{-u-v}}$