If u = x^2 tan^{-1}(frac{y}{x}) - y^2 tan^{-1 | vedclass

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

Question

(B) Given $u = x^2 	an^{-1}(\frac{y}{x}) - y^2 	an^{-1}(\frac{x}{y})$.Using the identity $	an^{-1}(\frac{x}{y}) = \frac{\pi}{2} - 	an^{-1}(\frac{y

Vedclass · Accepted Answer

$\frac{x^2 - y^2}{x^2 + y^2}$

Vedclass · Answer

(B) Given $u = x^2 	an^{-1}(\frac{y}{x}) - y^2 	an^{-1}(\frac{x}{y})$.Using the identity $	an^{-1}(\frac{x}{y}) = \frac{\pi}{2} - 	an^{-1}(\frac{y}{x})$ for $y > 0$,we have:$u = x^2 	an^{-1}(\frac{y}{x}) - y^2 (\frac{\pi}{2} - 	an^{-1}(\frac{y}{x}))$$u = (x^2 + y^2) 	an^{-1}(\frac{y}{x}) - \frac{\pi}{2} y^2$Now,differentiate $u$ partially with respect to $y$:$\frac{\partial u}{\partial y} = (x^2 + y^2) \cdot \frac{1}{1 + (y/x)^2} \cdot \frac{1}{x} + 2y 	an^{-1}(\frac{y}{x}) - \pi y$$\frac{\partial u}{\partial y} = (x^2 + y^2) \cdot \frac{x^2}{x^2 + y^2} \cdot \frac{1}{x} + 2y 	an^{-1}(\frac{y}{x}) - \pi y$$\frac{\partial u}{\partial y} = x + 2y 	an^{-1}(\frac{y}{x}) - \pi y$Next,differentiate $\frac{\partial u}{\partial y}$ partially with respect to $x$:$\frac{\partial^2 u}{\partial x \partial y} = 1 + 2y \cdot \frac{1}{1 + (y/x)^2} \cdot (-\frac{y}{x^2}) - 0$$\frac{\partial^2 u}{\partial x \partial y} = 1 + 2y \cdot \frac{x^2}{x^2 + y^2} \cdot (-\frac{y}{x^2})$$\frac{\partial^2 u}{\partial x \partial y} = 1 - \frac{2y^2}{x^2 + y^2} = \frac{x^2 + y^2 - 2y^2}{x^2 + y^2} = \frac{x^2 - y^2}{x^2 + y^2}$.

If $u = x^2 \tan^{-1}(\frac{y}{x}) - y^2 \tan^{-1}(\frac{x}{y})$,then $\frac{\partial^2 u}{\partial x \partial y} = $

Similar Questions

If $z = \frac{y}{x} \left[ \sin \left( \frac{x}{y} \right) + \cos \left( 1 + \frac{y}{x} \right) \right]$,then $x \frac{\partial z}{\partial x} = $

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

If $u = \log (x^3 + y^3 + z^3 - 3xyz)$,then $\left( \frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} + \frac{\partial u}{\partial z} \right) (x + y + z) =$

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module