Differentiation Question 238
Question: If $ u={{\tan }^{-1}}{ \frac{\sqrt{1+x^{2}}-1}{x} } $ and $ v=2{{\tan }^{-1}}x $ , then $ \frac{du}{dv} $ is equal to
[RPET 1997]
Options:
A) 4
B) 1
C) ¼
D) -1/4
Show Answer
Answer:
Correct Answer: C
Solution:
$ u={{\tan }^{-1}}{ \frac{\sqrt{1+x^{2}}-1}{x} } $ and $ v=2{{\tan }^{-1}}x $
Put $ x=\tan \theta $ in u and v; $ u={{\tan }^{-1}}{ \frac{\sqrt{1+{{\tan }^{2}}\theta }-1}{\tan \theta } } $ and $ v=2\theta $
$ u={{\tan }^{-1}}{ \frac{\sec \theta -1}{\tan \theta } } $ and $ v=2\theta $
$ \frac{3(x^{2}+y^{2}+z^{2}-xy-yz-zx)}{(x+y+z)(x^{2}+y^{2}+z^{2}-xy-yz-zx)} $ and $ v=2\theta $
$ u=\theta /2 $ and $ v=2\theta $ ;
$ \therefore \frac{du}{dv}=\frac{du/d\theta }{dv/d\theta }=\frac{1/2}{2}=\frac{1}{4} $ .
BETA
