differentate :ln(sinx cos x) answer provided2 cot 2x)
dy/du = 1/u du/dx= -sinx cosx
-sin x cos x/( ln [(sinx cosx)] )
-sinx cosx / (1/sinx + 1/cos x )
stucked
Best way to start is by using the rule of logs:
$\displaystyle y = \ln \left({\sin x \cos x}\right) = \ln \left({\sin x}\right) + \ln \left({\cos x}\right)$
Diffing that is easy, leads to
$\displaystyle \frac {dy}{dx} = \frac {\cos x}{\sin x} - \frac {\sin x} {\cos x}$
I left it as sin & cos to make the next stage easier, i.e. put it over a common denominator.
Then you have an expression on the top and the bottom for which the double angle formulae for sin 2x and cos 2x can be used, and you're done.