Question #169821

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


1
Expert's answer
2021-03-11T11:31:58-0500

y=3t2+2e3t+1t+2cos(3t)y=(t+2cos(3t))(6t+6e3t)(3t2+2e3t+1)(16sin(3t))(t+2cos(3t))2y=(3t2+(18t2+12e3t+6)sin(3t)(t+2cos(3t))2+6e3tt2e3t+(12t+12e3t)cos(3t)1)(t+2cos(3t))2\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}


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