Differentiation Question 60
Question: If $ y=\log \frac{1+\sqrt{x}}{1-\sqrt{x}}, $ then $ \frac{dy}{dx}= $
Options:
A) $ \frac{\sqrt{x}}{1-x} $
B) $ \frac{1}{\sqrt{x}(1-x)} $
C) $ \frac{\sqrt{x}}{1+x} $
D) $ \frac{1}{\sqrt{x}(1+x)} $
Show Answer
Answer:
Correct Answer: B
Solution:
$ t=\frac{5x+1}{10x^{2}-3}, $
Differentiating w.r.t. x of y, we get $ \frac{dy}{dx}=\frac{1-\sqrt{x}}{1+\sqrt{x}}[ \frac{(1-\sqrt{x})\frac{1}{2\sqrt{x}}+(1+\sqrt{x})\frac{1}{2\sqrt{x}}}{{{(1-\sqrt{x})}^{2}}} ] $
$ =\frac{1}{2(1-x)\sqrt{x}}[1-\sqrt{x}+1+\sqrt{x}]=\frac{1}{\sqrt{x}(1-x)} $ .
BETA
