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If $u=sin(x^2+y^2)$ and $a^2x^2+b^2y^2=c^2$ , find $\dfrac{\partial u}{\partial z}$
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u$=sin(x^2+y^2)$

Diff. partially w.r.t. x,
$\dfrac{\partial u}{\partial x} \;=\; cos(x^2+y^2) . (2x)$

Diff. partially w.r.t. y,
$\dfrac{\partial u}{\partial y} \;=\; cos(x^2+y^2) . (2y)$

Let, $f(x,y) \;=\; a^2x^2+b^2y^2-c^2=0$

$\therefore \dfrac{dy}{dx} \;=\; \dfrac{-\partial f / \partial x }{\partial f / \partial y} \;=\; \dfrac{-2a^2x}{2b^2y} \;=\; \dfrac{a^2x}{b^2y}$

We know that by chain rule,

$\dfrac{d u}{d x} \;=\; \dfrac{\partial u}{\partial x} . \dfrac{\partial x}{\partial x} + \dfrac{\partial u}{\partial y} . \dfrac{\partial y}{\partial x} \\ \; \\ \; \\ \therefore \dfrac{\partial u}{\partial x} \;=\; \dfrac{\partial u}{\partial x} . 1 + \dfrac{\partial u}{\partial y} . \dfrac{\partial y}{\partial x} \\ \; \\ \; \\ \therefore \dfrac{d u}{d x} \;=\; 2x. cos(x^2+y^2) + 2y . cos(x^2+y^2) \bigg( \dfrac{-a^2x}{b^2y} \bigg) \\ \; \\ \; \\ 2x. cos(x^2+y^2) + 2 . cos(x^2+y^2) \bigg( \dfrac{-a^2x}{b^2} \bigg) \\ \; \\ \; \\ \therefore \dfrac{d u}{d x} \;=\; 2x cos(x^2+y^2) \bigg[1- \dfrac{a^2x}{b^2} \bigg]$