If z = sec^{-1}left(frac{x^4+y^4-8x^2y^2}{x^2 | vedclass

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(B) Let $u = \sec z = \frac{x^4+y^4-8x^2y^2}{x^2+y^2}$.Here,$u$ is a homogeneous function of $x$ and $y$ with degree $n = \frac{4-2}{1} = 2$.By Euler'

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$2 \cot z$

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(B) Let $u = \sec z = \frac{x^4+y^4-8x^2y^2}{x^2+y^2}$.Here,$u$ is a homogeneous function of $x$ and $y$ with degree $n = \frac{4-2}{1} = 2$.By Euler's Theorem for homogeneous functions,$x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = n \cdot u$.Since $u = \sec z$,we have $\frac{\partial u}{\partial x} = \sec z 	an z \frac{\partial z}{\partial x}$ and $\frac{\partial u}{\partial y} = \sec z 	an z \frac{\partial z}{\partial y}$.Substituting these into Euler's theorem:$x (\sec z 	an z \frac{\partial z}{\partial x}) + y (\sec z 	an z \frac{\partial z}{\partial y}) = 2 \sec z$.Dividing both sides by $\sec z 	an z$:$x \frac{\partial z}{\partial x} + y \frac{\partial z}{\partial y} = \frac{2 \sec z}{\sec z 	an z} = 2 \cot z$.

If $z = \sec^{-1}\left(\frac{x^4+y^4-8x^2y^2}{x^2+y^2}\right)$,then $x \frac{\partial z}{\partial x} + y \frac{\partial z}{\partial y}$ is equal to

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