Question

If $ y=\log^{n}x $ , where $ \log ^{n} $ means $ \log \log \log ... $ (repeated $ n $ times), then $ x\log x \log ^{2}x\log ^{3}x...\log ^{n-1}x\log ^{n}x\frac{dy}{dx} $ is equal to:

• A. $ \log x $
• B. $ x $
• C. $ \frac{1}{\log x} $
• D. $ \log ^{n}x $

Answer 1

Answer: (D)

Given that, $ y=\log ^{n}x $
$ \therefore x\log x\log ^{2}x\log ^{3}x...\log ^{n-1}x\log ^{n}x\times \frac{dy}{dx} $
$ =x\log x\log^{2}x\log ^{3}x...\log ^{n-1}x\log^{n}x \times \frac{1}{x\log x\log^{2}x\log^{3}x...\log^{n-1}x} $
$ =\log^{n}x $