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

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