Question

Differentiate the following w.r.t. $x$:
$\frac{cos\,x}{log\,x},x>0$

Answer 1

The correct answer is $=\frac{-[xlog\,x.sin\,x+cos\,x]}{x(log\,x)^2},x>0$
Let $y=\frac{cos\,x}{log\,x}$
By using the chain rule,we obtain
$\frac{dy}{dx}=\frac{\frac{d}{dx}(cos\,x)\times log\,x-cos\,x\times \frac{d}{dx}(log\,x)}{(log\,x)^2}$
$=\frac{-sin\,xlog\,x-cos\,x\times \frac{1}{x}}{(log\,x)^2}$
$=\frac{-[xlog\,x.sin\,x+cos\,x]}{x(log\,x)^2},x>0$