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

Question

Dr. Abubacker · Accepted Answer

(b)
Given, $\sec{z = \frac{x^{4} + y^{4} - 8x^{2}y^{2}}{x^{2} + y^{2}}}$
Here, $n = 2$
$	herefore x\frac{\partial z}{\partial x} + y\frac{\partial z}{\partial y} = 2.\frac{\sec z}{\sec{z.	an z}} = 2\cot z$

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

Explanation

Reviewed by Dr. Abubacker

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