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

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(B) Let $f(x, y) = \sec z = \frac{x^4+y^4-8x^2y^2}{x^2+y^2}$.Since $f(x, y)$ is a homogeneous function of degree $n = 4 - 2 = 2$, by Euler's Theorem f

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$2 \cot z$

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(B) Let $f(x, y) = \sec z = \frac{x^4+y^4-8x^2y^2}{x^2+y^2}$.Since $f(x, y)$ is a homogeneous function of degree $n = 4 - 2 = 2$, by Euler's Theorem for homogeneous functions, we have $x \frac{\partial f}{\partial x} + y \frac{\partial f}{\partial y} = n f(x, y) = 2 \sec z$.Now, differentiating $f = \sec z$ with respect to $x$ and $y$, we get $\frac{\partial f}{\partial x} = \sec z 	an z \frac{\partial z}{\partial x}$ and $\frac{\partial f}{\partial y} = \sec z 	an z \frac{\partial z}{\partial y}$.Substituting these into Euler's equation:$x (\sec z 	an z \frac{\partial z}{\partial x}) + y (\sec z 	an z \frac{\partial z}{\partial y}) = 2 \sec z$.Dividing both sides by $\sec z 	an z$, we get:$x \frac{\partial z}{\partial x} + y \frac{\partial z}{\partial y} = \frac{2 \sec z}{\sec z 	an z} = 2 \cot z$.

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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