Question

If $u={\tan }^{-1}\left(\frac{x^3+y^3}{x-y}\right)$, then$x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}$is equal to

• A. $-\sin 2u$
• B. $\sin 2u$
• C. $\cos 2u$
• D. $-\cos 2u$

Answer 1

The correct option is B

$\sin 2u$

Explanation for correct option:

Finding the value of the expression:

Given equation,

$u={\tan }^{-1}\raisebox{1ex}{$\left(x^3+y^3\right)$}\!\left/ \!\raisebox{-1ex}{$\left(x-y\right)$}\right.$

so,

$\tan u=\raisebox{1ex}{$\left(x^3+y^3\right)$}\!\left/ \!\raisebox{-1ex}{$\left(x-y\right)$}\right.$

Let

$\tan u=z...\left(1\right)$

Therefore,

$z=\raisebox{1ex}{$\left(x^3+y^3\right)$}\!\left/ \!\raisebox{-1ex}{$\left(x-y\right)$}\right.$

Further simplifying the above equation,

$$\begin{gathered} z=\raisebox{1ex}{$x^3\left(1+\raisebox{1ex}{$y^3$}\!\left/ \!\raisebox{-1ex}{$x^3$}\right.\right)$}\!\left/ \!\raisebox{-1ex}{$x\left(1-\raisebox{1ex}{$y$}\!\left/ \!\raisebox{-1ex}{$x$}\right.\right)$}\right. \\ z=\raisebox{1ex}{$x^2\left(1+{\displaystyle \raisebox{1ex}{$y^3$}\!\left/ \!\raisebox{-1ex}{$x^3$}\right.}\right)$}\!\left/ \!\raisebox{-1ex}{$\left(1-{\displaystyle \raisebox{1ex}{$y$}\!\left/ \!\raisebox{-1ex}{$x$}\right.}\right)$}\right. \\ z=\raisebox{1ex}{$x^2\left(1+{\displaystyle {\left(\raisebox{1ex}{$y$}\!\left/ \!\raisebox{-1ex}{$x$}\right.\right)}^3}\right)$}\!\left/ \!\raisebox{-1ex}{$\left(1-{\displaystyle \raisebox{1ex}{$y$}\!\left/ \!\raisebox{-1ex}{$x$}\right.}\right)$}\right. \end{gathered}$$

From the above equation, it is clear that $z$ is a homogeneous function of the form$x^{n}f\left(\raisebox{1ex}{$y$}\!\left/ \!\raisebox{-1ex}{$x$}\right.\right).$

Here, $n=2$

We know, according to Euler's theorem,

$x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y}=nz...\left(2\right)$

Differentiate partially equation $\left(1\right)$ with respect to $x$ and $y$,

$\frac{\partial z}{\partial x}=sec2u\frac{\partial u}{\partial x}$

$\frac{\partial z}{\partial y}=sec2u\frac{\partial u}{\partial y}$

Putting the above values in equation $\left(2\right)$, we get,

$$\begin{gathered} xse{c}^{2}u\frac{\partial u}{\partial x}+yse{c}^{2}u\frac{\partial u}{\partial y}=2\tan u \\ \Rightarrow x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}=\raisebox{1ex}{$2\tan u$}\!\left/ \!\raisebox{-1ex}{$se{c}^{2}u$}\right. \\ \Rightarrow x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}=2\sin u\cos u \\ \Rightarrow x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}=\sin 2u \end{gathered}$$

Hence, the correct option is B.