Science:Math Exam Resources/Courses/MATH100/December 2011/Question 01 (k)/Solution 1

We compute this derivative using the chain rule twice and the product rule

${\begin{aligned}{\frac {dy}{dt}}&=e^{t\cos(2t)}\cdot {\frac {d}{dt}}(t\cos(2t))\\&=e^{t\cos(2t)}\cdot ({\frac {d}{dt}}(t)\cdot \cos(2t)+t{\frac {d}{dt}}(\cos(2t)))\\&=e^{t\cos(2t)}\cdot (1\cdot \cos(2t)+t(-\sin(2t)\cdot 2))\\&=e^{t\cos(2t)}\left(\cos(2t)-2t\sin(2t)\right)\end{aligned}}$