Differentiation Question 418
Question: If $ \cos x=\frac{1}{\sqrt{1+t^{2}}} $ and $ \sin y=\frac{t}{\sqrt{1+t^{2}}} $ , then $ \frac{dy}{dx}= $
[MP PET 1994]
Options:
A) -1
B) $ \frac{1-t}{1+t^{2}} $
C) $ \frac{1}{1+t^{2}} $
D) 1
Show Answer
Answer:
Correct Answer: D
Solution:
Obviously $ x={{\cos }^{-1}}\frac{1}{\sqrt{1+t^{2}}} $ and $ y={{\sin }^{-1}}\frac{t}{\sqrt{1+t^{2}}} $
Therefore $ \frac{dx}{dt}=-\frac{1}{\sqrt{\frac{t^{2}}{1+t^{2}}}}.\frac{(-1)}{2{{(1+t^{2})}^{3/2}}}2t=\frac{1}{1+t^{2}} $
and $ \frac{dy}{dt}=\sqrt{1+t^{2}}.\frac{1}{{{(\sqrt{1+t^{2}})}^{3/2}}}=\frac{1}{1+t^{2}} $
Hence $ \frac{dy}{dx}=\frac{dy/dt}{dx/dt}=1 $ .
BETA
