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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$

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  • 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 meena.p

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