Question

6. If u, v are function of r,s and r,s are functions of x,y then d(u,v) isa(x,y)/a(z.72) is​

Answer 1

SOLUTION

TO PROVE

If u, v are function of r,s and r,s are functions of x,y then

$\displaystyle \sf{ \frac{ \partial (u,v)}{ \partial (x,y)} =\frac{ \partial (u,v)}{ \partial (r,s)} . \frac{ \partial (r,s)}{ \partial (x,y)} }$

EVALUATION

Here it is given that u, v are function of r,s and r,s are functions of x,y

Now

RHS

$\displaystyle \sf{ = \frac{ \partial (u,v)}{ \partial (r,s)} . \frac{ \partial (r,s)}{ \partial (x,y)} }$

$\displaystyle = \begin{vmatrix} \frac{ \partial u}{ \partial r} & \frac{ \partial u}{ \partial s} \\ \\ \frac{ \partial v}{ \partial r} & \frac{ \partial v}{ \partial s} \end{vmatrix} \times \begin{vmatrix} \frac{ \partial r}{ \partial x} & \frac{ \partial s}{ \partial x} \\ \\ \frac{ \partial r}{ \partial y} & \frac{ \partial s}{ \partial y} \end{vmatrix}$

$\displaystyle = \sf{ \begin{vmatrix} \frac{ \partial u}{ \partial r}. \frac{ \partial r}{ \partial x} +\frac{ \partial u}{ \partial s}.\frac{ \partial s}{ \partial x} & \frac{ \partial u}{ \partial r}. \frac{ \partial r}{ \partial y} +\frac{ \partial u}{ \partial s}.\frac{ \partial s}{ \partial y} \\ \\ \frac{ \partial v}{ \partial r}. \frac{ \partial r}{ \partial x} +\frac{ \partial v}{ \partial s}.\frac{ \partial s}{ \partial x} & \frac{ \partial v}{ \partial r}. \frac{ \partial r}{ \partial y} +\frac{ \partial v}{ \partial s}.\frac{ \partial s}{ \partial y} \end{vmatrix} }$

$\displaystyle = \begin{vmatrix} \frac{ \partial u}{ \partial x} & \frac{ \partial u}{ \partial y} \\ \\ \frac{ \partial v}{ \partial x} & \frac{ \partial v}{ \partial y} \end{vmatrix}$

$\displaystyle \sf{ = \frac{ \partial (u,v)}{ \partial (x,y)}}$

= LHS

Hence proved

Additional Information :

Above rule is called Chain rule for Jacobians

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