# How do you find the derivative of lnx/cscx?

$\frac{1}{x} \sin x + \ln x \cos x$
$\left(\ln \frac{x}{\csc} x\right) ' = \left(\ln x \cdot \sin x\right) '$ so we use the product rule
$\left(\ln x \cdot \sin x\right) ' = \frac{1}{x} \sin x + \ln x \cos x$