If u = tan^{-1}(frac{y}{x}),then by Euler's T | vedclass

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(A) Given $u = 	an^{-1}(\frac{y}{x})$.We can rewrite this as $	an u = \frac{y}{x}$.Let $f(u) = 	an u = \frac{y}{x}$.Here,$f(u)$ is a homogeneous fu

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(A) Given $u = 	an^{-1}(\frac{y}{x})$.We can rewrite this as $	an u = \frac{y}{x}$.Let $f(u) = 	an u = \frac{y}{x}$.Here,$f(u)$ is a homogeneous function of $x$ and $y$ with degree $n = 0$ because $\frac{ty}{tx} = (\frac{y}{x})^0 \cdot \frac{y}{x}$.According to Euler's Theorem for homogeneous functions,if $f(u)$ is a homogeneous function of degree $n$,then $x \frac{\partial f}{\partial x} + y \frac{\partial f}{\partial y} = n f(u)$.Here,$f(u) = 	an u$ and $n = 0$.So,$x \frac{\partial (	an u)}{\partial x} + y \frac{\partial (	an u)}{\partial y} = 0 \cdot 	an u = 0$.Using the chain rule,$x \sec^2 u \frac{\partial u}{\partial x} + y \sec^2 u \frac{\partial u}{\partial y} = 0$.Dividing by $\sec^2 u$,we get $x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = 0$.

If $u = \tan^{-1}(\frac{y}{x})$,then by Euler's Theorem,the value of $x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y}$ is:

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