Answer to Question #169821 in Trigonometry for [e-mail]

Question

y=3t^2+2e^3t+1/t+2cos3t

Answer 1

$\displaystyle y = \frac{3t^2+2e^{3t}+1}{t+2\cos(3t)} \\ y' = \frac{\left(t+2\cos(3t)\right)\left(6t + 6e^{3t}\right) - \left(3t^2+2e^{3t}+1\right)\left(1 - 6\sin(3t)\right)}{\left(t+2\cos(3t)\right)^2} \\ y' = \frac{(3t^2 + (18t^2 + 12 e^{3t} + 6)\sin(3t)}{(t + 2\cos(3t))^2} + \\ \frac{6 e^{3t}t - 2 e^{3t} + (12t + 12 e^{3t})\cos(3t) - 1)}{(t + 2\cos(3t))^2}$