Use Quotient Rule $\displaystyle \frac{{dy}}{{dx}}=\frac{{v\frac{{du}}{{dx}}-u\frac{{dv}}{{dx}}}}{{{{v}^{2}}}}$

Differentiate the variables separately before inserting into the rule, there is less chance of making mistakes.

$\displaystyle u=\sin 2x\ \ \frac{{du}}{{dx}}=2\cos 2x\quad v=\ln {{x}^{2}}=2\ln x\quad \frac{{dv}}{{dx}}=\frac{2}{x}$

$\displaystyle \frac{{dy}}{{dx}}=\frac{{2\ln x\times 2\cos 2x-\sin 2x\times \frac{2}{x}}}{{{{{\left( {2\ln x} \right)}}^{2}}}}$

Multiply both the numerator and denominator by x to get rid of the x in  2/x.

$\displaystyle \frac{{dy}}{{dx}}=\frac{{x\times (2\ln x\times 2\cos 2x-\sin 2x\times \frac{2}{x})}}{{x\times {{{\left( {2\ln x} \right)}}^{2}}}}$

$\displaystyle \frac{{dy}}{{dx}}=\frac{{2x\ln x\times 2\cos 2x-2\sin 2x}}{{x\times {{{\left( {2\ln x} \right)}}^{2}}}}$

$\displaystyle \frac{{dy}}{{dx}}=\frac{{4x\cos 2x\ln x-2\sin 2x}}{{x\times {{2}^{2}}{{{\left( {\ln x} \right)}}^{2}}}}$

Factorize out 2 and divide it against the 2 in the denominator.

$\displaystyle \frac{{dy}}{{dx}}=\frac{{2(2x\cos 2x\ln x-\sin 2x)}}{{4x{{{\left( {\ln x} \right)}}^{2}}}}$

$\displaystyle \frac{{dy}}{{dx}}=\frac{{2x\cos 2x\ln x-\sin 2x}}{{2x{{{\left( {\ln x} \right)}}^{2}}}}$

Using Laws of Logarithm, the cos 2x in the expression cos 2x ln x will be raised  to be the power of lnx.

$\displaystyle \frac{{dy}}{{dx}}=\frac{{2x\ln {{x}^{{\cos 2x}}}-\sin 2x}}{{2x{{{(\ln x)}}^{2}}}}$

Additional Math – Differentiation – Sin 2x and 2 lnx. Additional Math Tuition. Woodlands, Choa Chu Kang, Yew Tee, Sembawang and Johor Bahru