Differentiation Question 431
Question: If $ \tan y=\frac{2t}{1-t^{2}} $ and $ \sin x=\frac{2t}{1+t^{2}}, $ then $ \frac{dy}{dx}= $
Options:
A) $ \frac{2}{1+t^{2}} $
B) $ \frac{1}{1+t^{2}} $
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Answer:
Correct Answer: C
Solution:
$ \tan y=\frac{2t}{1-t^{2}} $
…..(i) and $ \sin x=\frac{2t}{1+t^{2}} $
From (i), differentiating w.r.t. t of y, we get, $ \frac{dy}{dx}=\frac{1}{2}\frac{\cos (\sqrt{\sin x+\cos x})}{\sqrt{\sin x+\cos x}}(\cos x-\sin x) $ and $ \frac{dy}{dt}=\frac{2(1+t^{2})}{{{(1-t^{2})}^{2}}}.\frac{1}{(1+{{\tan }^{2}}y)} $
or $ \frac{dy}{dt}=\frac{2(1+t^{2})}{(1-t^{2})^{2}}\cdot\frac{1}{1+\left( \frac{2t}{1-t^{2}} \right)^{2}}=\frac{2}{1+t^{2}} $
…..(iii) and from (ii), differentiating w.r.t. t of x, we get $ \cos x\frac{dx}{dt}=\frac{2(1-t^{2})}{{{(1+t^{2})}^{2}}} $
or $ \frac{dx}{dt}=\frac{2(1-t^{2})}{{{(1+t^{2})}^{2}}}\frac{1}{\sqrt{1-\frac{{{(2t)}^{2}}}{{{(1+t^{2})}^{2}}}}}=\frac{2}{1+t^{2}} $
Hence $ \frac{dy}{dx}=0 $ .
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