If u = tan^{-1}(x + y), then xfrac{partial u} | vedclass

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(B) Given $u = 	an^{-1}(x + y)$,which implies $	an u = x + y$.Let $f(x, y) = 	an u = x + y$.Since $f(x, y)$ is a homogeneous function of degree $n

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$\frac{1}{2}\sin 2u$

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(B) Given $u = 	an^{-1}(x + y)$,which implies $	an u = x + y$.Let $f(x, y) = 	an u = x + y$.Since $f(x, y)$ is a homogeneous function of degree $n = 1$ in $x$ and $y$,by Euler's Theorem for homogeneous functions,we have:$x\frac{\partial f}{\partial x} + y\frac{\partial f}{\partial y} = n \cdot f(x, y)$$x\frac{\partial}{\partial x}(	an u) + y\frac{\partial}{\partial y}(	an u) = 1 \cdot 	an u$$x(\sec^2 u)\frac{\partial u}{\partial x} + y(\sec^2 u)\frac{\partial u}{\partial y} = 	an u$Dividing both sides by $\sec^2 u$,we get:$x\frac{\partial u}{\partial x} + y\frac{\partial u}{\partial y} = \frac{	an u}{\sec^2 u}$$x\frac{\partial u}{\partial x} + y\frac{\partial u}{\partial y} = \frac{\sin u}{\cos u} \cdot \cos^2 u = \sin u \cos u$Using the identity $\sin 2u = 2 \sin u \cos u$,we have $\sin u \cos u = \frac{1}{2} \sin 2u$.Therefore,$x\frac{\partial u}{\partial x} + y\frac{\partial u}{\partial y} = \frac{1}{2} \sin 2u$.

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

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