# Using Euler's theorem prove the following: if$\;u=\tan^{-1}\large[\frac{x^{3}+y^{3}}{x-y}]$ prove that $x\large\frac{\partial u}{\partial x}$+$y\large\frac{\partial u}{\partial y}=$$\sin 2u This question has multiple parts. Therefore each part has been answered as a separate question on Clay6.com ## 1 Answer Toolbox: • Euler's Theorem: If f(x,y) is a homogeneous function of degree n, then x\large\frac{\partial f}{\partial x}$$+y\large\frac{\partial f}{\partial y}$$=nf This can be extended to several variables • Chain rule for partial derivatives: • If w=f(u,v),u=g(x,y),v=u(x,y) then, • \large\frac{\partial w}{\partial x}=\frac{\partial w}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial w}{\partial v}\frac{\partial v}{\partial x},\frac{\partial w}{\partial y}=\frac{\partial w}{\partial u}\frac{\partial u}{\partial y}+\frac{\partial w}{\partial v}\frac{\partial v}{\partial y} u=\tan ^{-1} \bigg(\large\frac{x^3+y^3}{x-y}\bigg) Step 1: => \tan u= \large\frac{x^3+y^3}{x-y} Step 2: Let f(x,y) =\large\frac{x^3+y^3}{x-y}$$=\tan u$
F is a homogeneous function of degree 2 in x,y
$\therefore$ by Euler's theorem on homogeneous functions,
$x \large\frac{\partial f}{\partial x}$$+y \large\frac{\partial f}{\partial y}$$=2f$ -----(1)
Step 3:
$\therefore$ Now $f(x,y)=\tan u$
$\large\frac{\partial f}{\partial x}$$=sec^2 u \large\frac{\partial u}{\partial x} \large\frac{\partial f}{\partial y}$$=sec^2 u \large\frac{\partial u}{\partial y}$
and $3f=3 \tan u$
$\therefore$ (1) becomes
Step 4:
$x \sec^2u \large\frac{\partial u}{\partial x}$$+y\;sec^2 u \large\frac{\partial u}{\partial y}$$=2 \tan u \qquad (\div \sec^2x)$
$x \large\frac{\partial u}{\partial x}$$+y \large\frac{\partial u}{\partial y}$$=2 \large\frac{\sin u}{\cos u}$$\cos ^2 u (Since \large\frac{1}{\sec^2 x}$$=\cos ^2 x)$
$x \large\frac{\partial u}{\partial x}$$+y \large\frac{\partial u}{\partial y}$$=2 \sin u \cos u$
$x \large\frac{\partial u}{\partial x}$$+y \large\frac{\partial u}{\partial y}$$=\sin 2 u$

answered Aug 13, 2013 by

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