Limits Continuity And Differentiability Question 95
Question: If $ y={{\log }^{n}}x $ , where $ {{\log }^{n}} $ means log log log… (repeated n time), then $ xlogxlogxlog^{2}xlog^{3}x $
$ ….{{\log }^{n-1}}x{{\log }^{n}}x\frac{dy}{dx} $ is equal to
Options:
A) $ \log x $
B) $ {{\log }^{n}}x $
C) $ \frac{1}{\log x} $
D) 1
Show Answer
Answer:
Correct Answer: B
Solution:
$ \because y={{\log }^{n}}x $ On differentiating w.r.t.x, we get $ x\log x{{\log }^{2}}x{{\log }^{3}}x….{{\log }^{n-1}}x{{\log }^{n}}x\frac{dy}{dx} $
$ =\frac{x\log x{{\log }^{2}}x{{\log }^{3}}x….{{\log }^{n-1}}x{{\log }^{n}}x.1}{x\log x{{\log }^{2}}x{{\log }^{3}}x….{{\log }^{n-1}}x} $
$ ={{\log }^{n}}x $
BETA
