If u = tan^{-1}left(frac{x^3 + y^3}{x - y}rig | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given $u = 	an^{-1}\left(\frac{x^3 + y^3}{x - y}ight)$.This implies $	an u = \frac{x^3 + y^3}{x - y}$.Let $f(x, y) = 	an u = \frac{x^3 + y^3}

Vedclass · Accepted Answer

$\sin 2u$

Vedclass · Answer

(A) Given $u = 	an^{-1}\left(\frac{x^3 + y^3}{x - y}ight)$.This implies $	an u = \frac{x^3 + y^3}{x - y}$.Let $f(x, y) = 	an u = \frac{x^3 + y^3}{x - y}$.Here,$f(x, y)$ is a homogeneous function of degree $n = 3 - 1 = 2$.According to Euler's theorem for homogeneous functions,$x\frac{\partial f}{\partial x} + y\frac{\partial f}{\partial y} = n f$.Substituting $f = 	an u$ and $n = 2$,we get $x\frac{\partial}{\partial x}(	an u) + y\frac{\partial}{\partial y}(	an u) = 2 	an u$.Using the chain rule,$x \sec^2 u \frac{\partial u}{\partial x} + y \sec^2 u \frac{\partial u}{\partial y} = 2 	an u$.Dividing by $\sec^2 u$,we get $x\frac{\partial u}{\partial x} + y\frac{\partial u}{\partial y} = 2 \frac{	an u}{\sec^2 u}$.Since $\frac{	an u}{\sec^2 u} = \frac{\sin u / \cos u}{1 / \cos^2 u} = \sin u \cos u$,we have $2 \sin u \cos u = \sin 2u$.

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} = $

Similar Questions

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

$\begin{aligned} & f(x, y)=2(x-y)^2-x^4-y^4 \\ & \left|\left(f_{x x} f_{y y}-f_{x y}^2\right)\right|_{(0,0)} \end{aligned}$

If $f(x, y) = \frac{\cos(x - 4y)}{\cos(x + 4y)}$, then $\left. \frac{\partial f}{\partial x} \right|_{y = \frac{x}{4}}$ is equal to:

If $u=f(r)$, where $r^2=x^2+y^2$, then $\left(\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}\right)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module