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Find dy/dx when: x= sin2A y= cos4A
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Parametric equation?
Yes it is, sorry, I forgot to mention that!
y=cos4A dy=-4sin(4A) dA dy/dA = -4sin(4A) x=sin2A dx=2cos(2A) dA dx/dA = 2cos(2A) Use chain rule, dy/dA = (dy/dx)(dx/dA) dy/dx= (dy/dA)/(dx/dA) dy/dx = -2sin(4A)/cos(2A)
dunno if it works
Thank you all so much! :) I have a test later on and trig is a weak point ;) thanks for clearing that up though, much appreciated.
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@Callisto, the answer I have with the worksheet is -4sin2A.. thanks though, i've got it now :)
y = cos4a --> y' = -4sin4a = -8sin2acos2a x = sin2a ---> x' = 2cos2a -> y'/x' = -4sin2a
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