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Find $\large\frac{\partial w}{\partial u}$ and $\large\frac{\partial w}{\partial v}$ if$\;w=x^{2}+y^{2}$ where $x=u^{2}-v^{2},y=2uv$

This question has multiple parts. Therefore each part has been answered as a separate question on Clay6.com

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Toolbox:
  • Chain rule for partial derivatives:
  • If $w=f(u,v),u=g(x,y),v=u(x,y)$ then,
  • $\large\frac{\partial w}{\partial x}=\frac{\partial w}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial w}{\partial v}\frac{\partial v}{\partial x},\frac{\partial w}{\partial y}=\frac{\partial w}{\partial u}\frac{\partial u}{\partial y}+\frac{\partial w}{\partial v}\frac{\partial v}{\partial y}$
$w=x^2+y^2,\;x=u^2-v^2,\;y=2uv$
Step 1:
$\large\frac{\partial w}{\partial u}=\frac{\partial w}{\partial x}\frac{\partial x}{\partial u}+\frac{\partial w}{\partial y}\frac{\partial y}{\partial u}$
$\qquad=2x.2u+2y.2v$
$\qquad=4u(u^2-v^2)+4v.2uv$
$\qquad=4u(u^2-v^2+2v^2)$
$\qquad=4u(u^2+v^2)$
Step 2:
$\large\frac{\partial w}{\partial v}=\frac{\partial w}{\partial x}\frac{\partial x}{\partial v}+\frac{\partial w}{\partial y}\frac{\partial y}{\partial v}$
$\qquad=2x.(-2v)+2y.2u$
$\qquad=-4(u^2-v^2)v+4.2uv.u$
$\qquad=4v(-u^2+v^2+2u^2)$
$\qquad=4v(u^2+v^2)$
answered Aug 13, 2013 by meena.p
 

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