I needed to determine the derivative of e^3x(cos4x+4sin4x)
so far i have
f'(x)= 3e^3x(cos4x+4sin4x) + e^3x(16cos4x-4sin4x)
How can i simplfy it so i end up with an answer of
f'(x)= e^3x(19cos4x+8sin4x)<--- this is the final answer...I need the steps in between!
2007-12-03 15:24:53 · 5 answers · asked by michelle 1 in Science & Mathematics ➔ Mathematics
f (x) = e^(3x) cos(4x) + 4 e^(3x) sin(4x)
f `(x) is given by:-
A + B where:-
A = 3e^(3x) cos(4x) - 4sin(4x) e^(3x)
B = 12 e^(3x) sin(4x) + 16 e^(3x) cos(4x)
A + B is:-
19 e^(3x) cos (4x) + 8 e^(3x)sin(4x)
e^(3x) (19 cos 4x + 8 sin 4x)
2007-12-03 22:22:15 · answer #1 · answered by Como 7 · 2⤊ 0⤋
Well all you need to do is pull out e^3x. distribute the 3 and the combine your sins4x and cos4x like this
e^3x[3(cos4x+sin4x)+(16cos4x-4sin4x)]
e^3x[3cos4x+3sin4x+16cos4x-4sin4x)]
e^3x[19cos4x+8sin4x]
2007-12-03 15:33:12 · answer #2 · answered by jkinco 2 · 0⤊ 0⤋
There's not much step.
Multiply by the 3 factor (of the first term).
Factor out the common factor of e^3x from each of the 2 terms.
Combine similar terms in the other factor.
2007-12-03 15:34:37 · answer #3 · answered by answerING 6 · 0⤊ 0⤋
f(x) = e^3x(cos4x+4sin4x)
f'(x) = e^3x(-4xsin4x + 16cos4x) + 3e^3x(cos4x+4sin4x)
f'(x) = -4e^3x (sin4x) + 16e^3x(cos4x) + 3e^3x (cos4x) +12e^3x(sin4x)
f'(x) = 16e^3x(cos4x) + 3e^3x (cos4x) - 4e^3x (sin4x) )+ 12e^3x(sin4x)
f'(x) = 19e^3x(cos4x) + 8e^3x(sin4x)
f'(x) = e^3x (19cos4x + 8sinx4)
2007-12-03 15:37:39 · answer #4 · answered by ignoramus_the_great 7 · 0⤊ 0⤋
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2016-05-28 02:23:38 · answer #5 · answered by doris 3 · 0⤊ 0⤋