If u=x y^2 ^ -1 ((y)/(x)), then x (∂ u)/(∂ x)+ y (∂ u)/(∂ y) is equal to

Correct answer: C. 3 u Given that, aligned & u=x y^2 ^ -1 ((y)/(x)) & (∂ u)/(∂ x)=y^2 ^ -1 ((y)/(x))+x y^2 · (1)/(1+(y^2)/(x^2)) ·(-(y)/(x^2)) &=y^2 ^ -1 ((y)/(x))-(x y^3)/(x^2+y^2) & (∂ u)/(∂ y)=2 x y ^ -1 ((y)/(x))+x y^2 · (1)/(1+(y^2)/(x^2)) ·((1)/(x)) &=2 x y ^ -1 ((y)/(x))+(x^2 y^2)/(x^2+y^2) & x (∂ u)/(∂ x)+ y (∂ u)/(∂ y)=x y^2 ^ -1 ((y)/(x))-(x^2 y^3)/(x^2+y^2) & x (∂ u)/(∂ x)+y (∂ u)/(∂ y)=3 x y^2 ^ -1 (y)/(x)=3 u aligned

JEE Main

Exam: JEE Main
Subject: Mathematics
Chapter: Differentiation

JEE Main · Mathematics · Differentiation

If u=x y^2 ^ -1 ((y)/(x)), then x (∂ u)/(∂ x)+ y (∂ u)/(∂ y) is...

u=x y^2 \tan ^{-1}

\left(\frac{y}{x}\right), then

x

\frac{\partial u}{\partial x}+

y

\frac{\partial u}{\partial y} is equal to

Options

A $2 u$
B $u$
C $3 u$
D $\frac{1}{3} u$

Correct answer

C. $3 u$

Step-by-step solution

Given that,

\begin{aligned} & u=x y^2 \tan ^{-1}\left(\frac{y}{x}\right) \\ & \frac{\partial u}{\partial x}=y^2 \tan ^{-1}\left(\frac{y}{x}\right)+x y^2 \cdot \frac{1}{1+\frac{y^2}{x^2}} \cdot\left(-\frac{y}{x^2}\right) \\ &=y^2 \tan ^{-1}\left(\frac{y}{x}\right)-\frac{x y^3}{x^2+y^2} \\ & \frac{\partial u}{\partial y}=2 x y \tan ^{-1}\left(\frac{y}{x}\right)+x y^2 \cdot \frac{1}{1+\frac{y^2}{x^2}} \cdot\left(\frac{1}{x}\right) \\ &=2 x y \tan ^{-1}\left(\frac{y}{x}\right)+\frac{x^2 y^2}{x^2+y^2} \\ & \therefore \quad x \frac{\partial u}{\partial x}+ y \frac{\partial u}{\partial y}=x y^2 \tan ^{-1}\left(\frac{y}{x}\right)-\frac{x^2 y^3}{x^2+y^2} \\ & x \frac{\partial u}{\partial x}+y \frac{\partial u}{\partial y}=3 x y^2 \tan ^{-1} \frac{y}{x}=3 u\end{aligned}

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