[tex]y = \ln(x) \cos(2x)[/tex]
Take the logarithm of both sides.
[tex]\ln(y) = \ln\bigg(\ln(x) \cos(2x)\bigg)[/tex]
Expand the right side.
[tex]\ln(y) = \ln\bigg(\ln(x)\bigg) + \ln\bigg(\cos(2x)\bigg)[/tex]
Use the chain rule to differentiate both sides with respect to [tex]x[/tex].
[tex]\dfrac{y'}y = \dfrac{\bigg(\ln(x)\bigg)'}{\ln(x)} + \dfrac{\bigg(\cos(2x)\bigg)'}{\cos(2x)}[/tex]
[tex]\dfrac{y'}y = \dfrac1{x\ln(x)} - \dfrac{\sin(2x)\bigg(2x\bigg)'}{\cos(2x)}[/tex]
[tex]\dfrac{y'}y = \dfrac1{x\ln(x)} - 2 \tan(2x)[/tex]
Solve for [tex]y'[/tex].
[tex]\dfrac{y'}y = \dfrac{1 - 2x \ln(x) \tan(2x)}{x\ln(x)}[/tex]
[tex]y' = \dfrac{1 - 2x \ln(x) \tan(2x)}{x\ln(x)}\,y[/tex]
[tex]y' = \dfrac{1 - 2x \ln(x) \tan(2x)}{x\ln(x)}\,\bigg(\ln(x) \cos(2x)\bigg)[/tex]
[tex]y' = \boxed{\dfrac{\cos(2x) - 2x \ln(x) \sin(2x)}x}[/tex]
