Differentiation Question 442
Question: If $ x=\frac{1-t^{2}}{1+t^{2}} $ and $ y=\frac{2t}{1+t^{2}} $ , then $ \frac{dy}{dx}= $
[Karnataka CET 2000; Pb. CET 2002]
Options:
A) $ \frac{-y}{x} $
B) $ \frac{y}{x} $
C) $ \frac{-x}{y} $
D) $ \frac{x}{y} $
Show Answer
Answer:
Correct Answer: C
Solution:
$ x=\frac{1-t^{2}}{1+t^{2}} $ and $ y=\frac{2t}{1+t^{2}} $
Put $ t=\tan \theta $ in both the equations, we get $ x=\frac{1-{{\tan }^{2}}\theta }{1+{{\tan }^{2}}\theta }=\cos 2\theta $ and $ y=\frac{2\tan \theta }{1+{{\tan }^{2}}\theta }=\sin 2\theta $ . Differentiating both the equations, we get $ \frac{dx}{d\theta }=-2\sin 2\theta $ and $ \frac{dy}{d\theta }=2\cos 2\theta . $
Therefore $ \frac{dy}{dx}=-\frac{\cos 2\theta }{\sin 2\theta }=-\frac{x}{y} $ .
BETA
