Differentiation Question 307
Question: If $ y=\sec ({{\tan }^{-1}}x), $ then $ \frac{dy}{dx} $ is
[DCE 2002; Kurukshetra CEE 2001]
Options:
A) $ \frac{x}{\sqrt{1+x^{2}}} $
B) $ \frac{-x}{\sqrt{1+x^{2}}} $
C) $ \frac{x}{\sqrt{1-x^{2}}} $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
$ y=\sec ({{\tan }^{-1}}x) $
$ \therefore \theta =\frac{1}{2}{{\cos }^{-1}}x $
$ =\frac{x}{1+x^{2}}.\sqrt{1+x^{2}}=\frac{x}{\sqrt{1+x^{2}}} $ , $ ({{\tan }^{-1}}x={{\sec }^{-1}}\sqrt{1+x^{2}}) $ .
BETA
