Differentiation Question 228
Question: If $ y=\sqrt{\frac{1+\tan x}{1-\tan x}} $ , then $ \frac{dy}{dx}= $
[AISSE 1981, 83, 84, 85; DSSE 1985; AI CBSE 1981, 83]
Options:
A) $ \frac{1}{2}\sqrt{\frac{1-\tan x}{1+\tan x}}.{{\sec }^{2}}( \frac{\pi }{4}+x ) $
B) $ \sqrt{\frac{1-\tan x}{1+\tan x}}.{{\sec }^{2}}( \frac{\pi }{4}+x ) $
C) $ \frac{1}{2}\sqrt{\frac{1-\tan x}{1+\tan x}}.\sec ( \frac{\pi }{4}+x ) $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
$ y=\sqrt{( \frac{1+\tan x}{1-\tan x} )} $ or $ y=\sqrt{\tan ( \frac{\pi }{4}+x )} $
$ \frac{dy}{dx}=\frac{1}{2\sqrt{\tan ( \frac{\pi }{4}+x )}}{{\sec }^{2}}( \frac{\pi }{4}+x ) $
$ =\frac{1}{2}\sqrt{[ \frac{1-\tan x}{1+\tan x} ]}{{\sec }^{2}}( \frac{\pi }{4}+x ) $ .
BETA
