If xy = tan^{-1}(xy) + cot^{-1}(xy),then left | vedclass

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

Question

(A) We know the trigonometric identity $	an^{-1}(u) + \cot^{-1}(u) = \frac{\pi}{2}$ for all $u \in \mathbb{R}$.Given the equation $xy = 	an^{-1}(xy)

Vedclass · Accepted Answer

$\frac{-1}{2}$

Vedclass · Answer

(A) We know the trigonometric identity $	an^{-1}(u) + \cot^{-1}(u) = \frac{\pi}{2}$ for all $u \in \mathbb{R}$.Given the equation $xy = 	an^{-1}(xy) + \cot^{-1}(xy)$,we can substitute the identity to get $xy = \frac{\pi}{2}$.Differentiating both sides with respect to $x$ using the product rule:$y + x \frac{dy}{dx} = 0$.Solving for $\frac{dy}{dx}$,we get $\frac{dy}{dx} = -\frac{y}{x}$.Evaluating this at the point $(4, 2)$:$\left(\frac{dy}{dx}ight)_{(4,2)} = -\frac{2}{4} = -\frac{1}{2}$.

If $xy = \tan^{-1}(xy) + \cot^{-1}(xy)$,then $\left(\frac{dy}{dx}\right)_{(4,2)} = ?$ (where $x, y \in \mathbb{R}$)

Similar Questions

If $x \sin (\alpha+y)=\sin y$ and $y=\frac{m}{x^2+2 n x+1}$ then $m^2=$

If ${x^2} + {y^2} = 1$,then find the relation between $y'$ and $y''$,where $y' = \frac{dy}{dx}$ and $y'' = \frac{d^2y}{dx^2}$.

If $y = \sqrt{\sin x + y},$ then $\frac{dy}{dx}$ is equal to

If $x^y = e^{x - y}$,then $\frac{dy}{dx}$ at $x = 1$ is . . . . . .

If ${x^{2/3}} + {y^{2/3}} = {a^{2/3}}$,then $\frac{dy}{dx} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module