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Find $\large\frac{\partial w}{\partial u}$ and $\large\frac{\partial w}{\partial v}$ if $\;w=\sin^{-1}xy$ where $x=u+v,y=u-v$

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,
  • $\large\frac{\partial w}{\partial x}=\frac{\partial w}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial w}{\partial v}\frac{\partial v}{\partial x},\frac{\partial w}{\partial y}=\frac{\partial w}{\partial u}\frac{\partial u}{\partial y}+\frac{\partial w}{\partial v}\frac{\partial v}{\partial y}$
$w=\sin ^{-1} xy\;x=u+v,\;y=u-v$
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=\large\frac{y}{\sqrt {1-x^2y^2}}$$(1) +\large\frac{x}{\sqrt {1-x^2y^2}}$$(1)$
$\qquad=\large\frac{u-v+u+v}{\sqrt {1-(u+v)^2(u-v)^2}}=\frac{2u}{\sqrt {1-(u^2-v^2)}}$
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=\large\frac{y}{\sqrt {1-x^2y^2}}$$(1) +\large\frac{x}{\sqrt {1-x^2y^2}}$$(-1)$
$\qquad=\large\frac{u-v-(u+v)}{\sqrt {1-(u+v)^2(u-v)^2}}=\frac{-2v}{\sqrt {1-(u^2-v^2)}}$
answered Aug 13, 2013 by meena.p
 

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