Question

Hi all! checking my answer q:"Differentiate y=e^-2t *cos4t"
I got:(-2e^-2t)(cos4t)+(-sin4t)(e^-2t)

Answer 1

i get
[cos(4t)-4tsin(4t) ] / e^2

Answer 2

or i guess it can be written as
(e^-2)cos(4t)-4tsin(4t)(e^-2)

Answer 3

hmm

Answer 4

here's what I got: f(x):e^-2t g(x):cos4t g'(x):-sin4t
for f'(x) I split the function into g(x):-2t and f(u):e^x which means g'(x)=(-2)(e^-2t)
so y': (-2e^-2t)(cos4t)+(-sin4t)(e^-2t)

Answer 5

oops the derivative of cos4t is -4sin(4t)

Answer 6

so, y': (-2e^-2t)(cos4t)+(-4sin4t)(e^-2t)

Answer 7

Well, I guess that's right