Differentiation Question 227
Question: If $ y=\sin ( \frac{1+x^{2}}{1-x^{2}} ) $ , then $ \frac{dy}{dx}= $
[AISSE 1987]
Options:
A) $ \frac{4x}{1-x^{2}}.\cos ( \frac{1+x^{2}}{1-x^{2}} ) $
B) $ \frac{x}{{{(1-x^{2})}^{2}}}.\cos ( \frac{1+x^{2}}{1-x^{2}} ) $
C) $ \frac{x}{(1-x^{2})}.\cos ( \frac{1+x^{2}}{1-x^{2}} ) $
D) $ \frac{4x}{{{(1-x^{2})}^{2}}}.\cos ( \frac{1+x^{2}}{1-x^{2}} ) $
Show Answer
Answer:
Correct Answer: D
Solution:
$ \frac{dy}{dx}=\cos ( \frac{1+x^{2}}{1-x^{2}} )[ \frac{(1-x^{2})2x+(1+x^{2})2x}{{{(1-x^{2})}^{2}}} ] $
$ =\frac{4x}{{{(1-x^{2})}^{2}}}\cos ( \frac{1+x^{2}}{1-x^{2}} ) $ .
BETA
