Please can you check my working out?
Differentiate the following with respect to x
\(\displaystyle \L y = e^{\sin 2x} \cos 2x\)
\(\displaystyle \L u = e^{\sin 2x} {\rm }u' = 2\cos 2x.e^{\sin 2x}\)
\(\displaystyle \L v = \cos 2x{\rm }v' = - 2\sin 2x\)
\(\displaystyle \L \frac{{dy}}{{dx}} = \cos 2x.2\cos 2x.e^{\sin 2x} - e^{\sin 2x} .2\sin 2x\)
\(\displaystyle \L = 2\cos ^2 2x.e^{\sin 2x} - e^{\sin 2x} .2\sin 2x\)
\(\displaystyle \L = 2e^{\sin 2x} \left( {\cos ^2 2x - \sin 2x} \right)\)
Thanks
Differentiate the following with respect to x
\(\displaystyle \L y = e^{\sin 2x} \cos 2x\)
\(\displaystyle \L u = e^{\sin 2x} {\rm }u' = 2\cos 2x.e^{\sin 2x}\)
\(\displaystyle \L v = \cos 2x{\rm }v' = - 2\sin 2x\)
\(\displaystyle \L \frac{{dy}}{{dx}} = \cos 2x.2\cos 2x.e^{\sin 2x} - e^{\sin 2x} .2\sin 2x\)
\(\displaystyle \L = 2\cos ^2 2x.e^{\sin 2x} - e^{\sin 2x} .2\sin 2x\)
\(\displaystyle \L = 2e^{\sin 2x} \left( {\cos ^2 2x - \sin 2x} \right)\)
Thanks
