Question

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

• A. $\cot z$
• B. $2\cot z$
• C. $2\tan z$
• D. $2\sec z$

Answer 1

The correct option is B

$2\cot z$

Explanation for the correct option.

Find the value of the expression.

The equation $z={\sec }^{-1}\left(\frac{x^4+y^4-8x^2{y}^2}{x^2+y^2}\right)$ can be written as:

$\sec z=\frac{x^4+y^4-8x^2{y}^2}{x^2+y^2}$.

Now, let $f\left(x,y\right)=\sec z$, where $f(x,y)=\frac{x^4+y^4-8x^2{y}^2}{x^2+y^2}$ is a homogenous equation of degree $2$.

Now it can be written that: $x\frac{\partial f}{\partial x}+y\frac{\partial f}{\partial y}=nf$ where $f=\sec z$ and $n$ is degree so $n=2$.

$$\begin{gathered} x\frac{\partial \sec z}{\partial x}+y\frac{\partial \sec z}{\partial y}=2\sec z \\ \Rightarrow x\sec z\tan z\frac{\partial z}{\partial x}+y\sec z\tan z\frac{\partial z}{\partial y}=2\sec z \\ \Rightarrow \sec z\tan z\left(x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y}\right)=2\sec z \\ \Rightarrow x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y}=\frac{2\sec z}{\sec z\tan z} \\ \Rightarrow x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y}=\frac{2}{\tan z} \\ \Rightarrow x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y}=2\cot z \end{gathered}$$

So the value of $x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y}$ is $2\cot z$.

Hence, the correct option is B.