Question

Differentiate the following w.r.t. $x$:
$log(log\,x),x>1$

Answer 1

The correct answer is $=\frac{1}{xlog\,x},x>1$
Let $y=log(log\,x)$
By using the chain rule,we obtain
$\frac{dy}{dx}=\frac{d}{dx}[log(log\,x)]$
$=\frac{1}{log\,x}.\frac{d}{dx}(log\,x)$
$=\frac{1}{log\,x}.\frac{1}{x}$
$=\frac{1}{xlog\,x},x>1$