If u = tan^{-1}left(frac{x^3 + y^3}{x - y}rig | vedclass

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(A) Given $u = 	an^{-1}\left(\frac{x^3 + y^3}{x - y}ight)$.This implies $	an u = \frac{x^3 + y^3}{x - y}$.Let $f(x, y) = 	an u = \frac{x^3 + y^3}

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$\sin 2u$

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(A) Given $u = 	an^{-1}\left(\frac{x^3 + y^3}{x - y}ight)$.This implies $	an u = \frac{x^3 + y^3}{x - y}$.Let $f(x, y) = 	an u = \frac{x^3 + y^3}{x - y}$.Here,$f(x, y)$ is a homogeneous function of degree $n = 3 - 1 = 2$.According to Euler's theorem for homogeneous functions,$x\frac{\partial f}{\partial x} + y\frac{\partial f}{\partial y} = n f$.Substituting $f = 	an u$ and $n = 2$,we get $x\frac{\partial}{\partial x}(	an u) + y\frac{\partial}{\partial y}(	an u) = 2 	an u$.Using the chain rule,$x \sec^2 u \frac{\partial u}{\partial x} + y \sec^2 u \frac{\partial u}{\partial y} = 2 	an u$.Dividing by $\sec^2 u$,we get $x\frac{\partial u}{\partial x} + y\frac{\partial u}{\partial y} = 2 \frac{	an u}{\sec^2 u}$.Since $\frac{	an u}{\sec^2 u} = \frac{\sin u / \cos u}{1 / \cos^2 u} = \sin u \cos u$,we have $2 \sin u \cos u = \sin 2u$.

If $u = \tan^{-1}\left(\frac{x^3 + y^3}{x - y}\right)$,then $x\frac{\partial u}{\partial x} + y\frac{\partial u}{\partial y} = $

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