Question

If $x=\frac{1-t^2}{1+t^2}$ and $y=\frac{2t}{1+t^2}$ at then $\frac{dy}{dx}=?$

• A. $\frac{-1}{x^{3}y}$
• B. $\frac{y}{x}$
• C. $\frac{-x}{y}$
• D. $\frac{x}{y}$

Answer 1

Answer: (A)

$\frac{-1}{x^{3}y}$

$x=\frac{1-t^2}{1+t^2}$ & $y=\frac{2t}{1+t^2}$

Let $t=\tan \theta$

$x=\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }$ $y=\frac{2\tan \theta }{1+{\tan }^2\theta }$

$x=\cos 2\theta$ $y=\tan 2\theta$

$\frac{dx}{d\theta }=-2\sin 2\theta$ $\frac{dy}{d\theta }=2{\sec }^{2}2\theta$

$\frac{dy}{dx}=\frac{dy}{dt}\times \frac{d\theta }{dx}=\frac{2{\sec }^{2}2\theta }{-2\sin 2\theta }=\frac{-{\sec }^{2}2\theta }{\sin 2\theta }$

$=-\frac{1}{{\cos }^{2}2\theta \sin 2\theta }\times \frac{\cos 2\theta }{\cos 2\theta }$

$=-\frac{1}{{\cos }^{3}2\theta \tan 2\theta }$

$=-\frac{1}{x^{3}y}$