Question

Вычислить производную функции:

Answer 1

$y=\sin x\cdot \ln 2x - 3arctg^2x$

$y' = (\sin x\cdot \ln 2x - 3arctg^2x)' = (\sin x\cdot \ln 2x)' - (3arctg^2x)' = (\sin x)'\ln 2x +\\\\+ \sin x(\ln 2x)' - 3\cdot 2\cdot arctgx\cdot \dfrac{1}{1+x^2} = \cos x\ln 2x + \sin x\cdot 2\cdot \dfrac{1}{2x} - \dfrac{6arctgx}{1+x^2} =\\\\\\= \boxed{\cos x\ln 2x + \dfrac{\sin x}{x} - \dfrac{6arctgx}{1+x^2}}$