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If $y=log_3 (log_3x), $ then $\frac{dy}{dx}=$ |
$\frac{1}{xlog3x}$ $\frac{1}{x(xlog3+logx)}$ $\frac{logx}{xlog3}$ $\frac{1}{xlogx\, log 3}$ |
$\frac{1}{xlogx\, log 3}$ |
The correct answer is Option (4) → $\frac{1}{xlogx\, log 3}$ $y=\log_3(\log_5x)$ $=\frac{ln(\log_3x)}{ln\,3}$ $⇒\frac{dy}{dx}=\frac{1}{ln\,3}\frac{d}{dx}[ln(\log_3x)]$ $=\frac{1}{ln\,3\log_3x.x}$ |
