Asked by Hannah
How would I differentiate
y = 4cos(2x)cos(3x)
u = cos(2x)
u' = -2sin(2x)
v = cos(3x)
v' = -3sin(3x)
This is all I have so far
y = 4cos(2x)cos(3x)
u = cos(2x)
u' = -2sin(2x)
v = cos(3x)
v' = -3sin(3x)
This is all I have so far
Answers
Answered by
Reiny
you have to use the product rule,
with your notation, it would be
d(uv) = uv' + vu'
= (-8sin(2x))(cos(3x)) + (4cos(2x)(-3sin(3x) )
= -8sin(2x)cos(3x) - 4sin(3x)cos(2x)
with your notation, it would be
d(uv) = uv' + vu'
= (-8sin(2x))(cos(3x)) + (4cos(2x)(-3sin(3x) )
= -8sin(2x)cos(3x) - 4sin(3x)cos(2x)
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