Find $\frac{\partial w}{\partial u}$ and $\frac{\partial w}{\partial v}$ in terms of $u$ and $v$ if $w = x^{2} + y^{2}$ where $x = u^{2} - v^{2}$ and $y = 2 u v$ , using chain rule.

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Solution

$w = x^{2} + y^{2}$ , $x = u^{2} - v^{2}$ , $y = 2 u v$
$\frac{\partial w}{\partial x} = 2 x$ $\frac{\partial x}{\partial u} = 2 u$ $\frac{\partial y}{\partial u} = 2 v$
$\frac{\partial w}{\partial y} = 2 y$ $\frac{\partial x}{\partial v} = - 2 v$ $\frac{\partial y}{\partial v} = 2 u$

$\frac{\partial ω}{\partial u} = \frac{\partial ω}{\partial x} \frac{\partial x}{\partial u} + \frac{\partial ω}{\partial u} \frac{\partial u}{\partial y}$
$= 2 x (2 u) + 2 y (2 v)$
$= 4 x u + 4 y v$
$= 4 (u^{2} - v^{2}) u + 4 (2 u v) \cdot v$
$= 4 [u^{3} + {u v}^{2}] = 4 u (u^{2} + v^{2})$

$\frac{\partial ω}{\partial v} = \frac{\partial ω}{\partial x} \frac{\partial x}{\partial v} + \frac{\partial ω}{\partial y} \frac{\partial y}{\partial v}$
$= 2 x (- 2 v) + 2 y (2 u)$
$= - 4 [(u^{2} - v^{2}) v - 2 u v \cdot u]$
$= - 4 (u^{2} v - v^{3} - 2 u^{2} v) = 4 v (u^{2} + v^{2})$

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\vec{u,} \vec{v} and \vec{w}

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If u, v and w are functions of x, then show that

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$w = x^{2} + y^{2}$ ,	$x = u^{2} - v^{2}$ ,	$y = 2 u v$
$\frac{\partial w}{\partial x} = 2 x$	$\frac{\partial x}{\partial u} = 2 u$	$\frac{\partial y}{\partial u} = 2 v$
$\frac{\partial w}{\partial y} = 2 y$	$\frac{\partial x}{\partial v} = - 2 v$	$\frac{\partial y}{\partial v} = 2 u$

Find ∂w∂u and ∂w∂v in terms of u and v if w=x2+y2 where x=u2−v2 and y=2uv, using chain rule.

Find $\frac{\partial w}{\partial u}$ and $\frac{\partial w}{\partial v}$ in terms of $u$ and $v$ if $w = x^{2} + y^{2}$ where $x = u^{2} - v^{2}$ and $y = 2 u v$ , using chain rule.