Question

y=log(tanx)×cotx by chain rule.​

Answer 1

$y = ln( \tan(x) ) \cot(x) \\ \frac{dy}{dx} = \frac{1}{ \tan(x) } \times {sec}^{2} x \times \cot(x) + ln( \tan(x) ) \times ( - {cosec}^{2} x$

hope it will help you!!