If y = x operatorname{Tan}^{-1}left(frac{x}{y | vedclass

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

Question

(C) Given the equation $y = x \operatorname{Tan}^{-1}\left(\frac{x}{y}\right)$.Divide by $x$: $\frac{y}{x} = \operatorname{Tan}^{-1}\left(\frac{x}{y}\

Vedclass · Accepted Answer

$\frac{y}{x}$

Vedclass · Answer

(C) Given the equation $y = x \operatorname{Tan}^{-1}\left(\frac{x}{y}ight)$.Divide by $x$: $\frac{y}{x} = \operatorname{Tan}^{-1}\left(\frac{x}{y}ight)$.Let $v = \frac{y}{x}$,so $y = vx$. Then $\frac{x}{y} = \frac{1}{v}$.The equation becomes $v = \operatorname{Tan}^{-1}\left(\frac{1}{v}ight)$,which implies $	an(v) = \frac{1}{v}$,or $v 	an(v) = 1$.However,differentiating $y = x \operatorname{Tan}^{-1}\left(\frac{x}{y}ight)$ with respect to $x$ using the product rule and chain rule:$\frac{dy}{dx} = \operatorname{Tan}^{-1}\left(\frac{x}{y}ight) + x \cdot \frac{1}{1 + (x/y)^2} \cdot \frac{d}{dx}\left(\frac{x}{y}ight)$$\frac{dy}{dx} = \operatorname{Tan}^{-1}\left(\frac{x}{y}ight) + x \cdot \frac{y^2}{x^2+y^2} \cdot \left(\frac{y - x \frac{dy}{dx}}{y^2}ight)$$\frac{dy}{dx} = \frac{y}{x} + \frac{x}{x^2+y^2} \cdot (y - x \frac{dy}{dx})$$\frac{dy}{dx} = \frac{y}{x} + \frac{xy}{x^2+y^2} - \frac{x^2}{x^2+y^2} \frac{dy}{dx}$$\frac{dy}{dx} \left(1 + \frac{x^2}{x^2+y^2}ight) = \frac{y}{x} + \frac{xy}{x^2+y^2}$$\frac{dy}{dx} \left(\frac{x^2+y^2+x^2}{x^2+y^2}ight) = \frac{y(x^2+y^2) + x^2y}{x(x^2+y^2)}$$\frac{dy}{dx} \left(\frac{2x^2+y^2}{x^2+y^2}ight) = \frac{yx^2+y^3+x^2y}{x(x^2+y^2)} = \frac{2x^2y+y^3}{x(x^2+y^2)}$$\frac{dy}{dx} = \frac{y(2x^2+y^2)}{x(x^2+y^2)} \cdot \frac{x^2+y^2}{2x^2+y^2} = \frac{y}{x}$.

If $y = x \operatorname{Tan}^{-1}\left(\frac{x}{y}\right)$,then $\frac{dy}{dx} = $

Similar Questions

If $\log (x+y)=2xy$,then $\frac{dy}{dx}$ at $x=0$ is

If ${2^x} + {2^y} = {2^{x + y}},$ then the value of $\frac{dy}{dx}$ at $x = y = 1$ is

If ${x^2} + {y^2} = t - \frac{1}{t}$ and ${x^4} + {y^4} = {t^2} + \frac{1}{{{t^2}}}$,then $\frac{{dy}}{{dx}}$ equals:

If $\log (x+y)=\sin (x+y)$,then $\frac{dy}{dx}$ is

If $y = \sqrt{(1 - x)(1 + x)}$,then

Vedclass Test Series

Exam Paper Generator

Online Exam Module