Differentiation Question 432
Question: If $ y=\sqrt{\frac{1+x}{1-x}}, $ then $ \frac{dy}{dx}= $
[AISSE 1981; RPET 1995]
Options:
A) $ \frac{2}{{{(1+x)}^{1/2}}{{(1-x)}^{3/2}}} $
B) $ \frac{1}{{{(1+x)}^{1/2}}{{(1-x)}^{3/2}}} $
C) $ \frac{1}{2{{(1+x)}^{1/2}}{{(1-x)}^{3/2}}} $
D) $ \frac{1}{{{(1+x)}^{3/2}}{{(1-x)}^{1/2}}} $
Show Answer
Answer:
Correct Answer: B
Solution:
$ y=\sqrt{\frac{1+x}{1-x}} $
Therefore $ y=\sqrt{\frac{(1+x)(1+x)}{(1-x)(1+x)}}=\sqrt{\frac{{{(1+x)}^{2}}}{1-x^{2}}} $
Differentiating with respect to x, we get $ \frac{dy}{dx}=\frac{{{(1-x)}^{1/2}}\frac{d}{dx}{{(1+x)}^{1/2}}-{{(1+x)}^{1/2}}\frac{d}{dx}{{(1-x)}^{1/2}}}{(1-x)} $
$ =\frac{(1-x)+(1+x)}{2{{(1-x)}^{3/2}}{{(1+x)}^{1/2}}} $
$ x=\frac{2}{3}y $ . Trick : $ \log y=\frac{1}{2}\log (1+x)-\frac{1}{2}\log (1-x) $
Therefore $ \frac{1}{y}\frac{dy}{dx}=\frac{1}{2(1+x)}+\frac{1}{2(1-x)} $
Therefore $ \frac{dy}{dx}=\frac{1}{(1+x)(1-x)}\times \sqrt{\frac{1+x}{1-x}} $
$ =\frac{1}{{{(1+x)}^{1/2}}{{(1-x)}^{3/2}}} $ .
BETA
