Differentiating 2cos2t (Parametic Equations) Watch

Stephhcharlene
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#1
Report Thread starter 7 years ago
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Hi there,

I have y=2cos2t and need to do the whole dy/dx etcetc. I know I can rearrange this to be 2(2cos^2t-1), 2(1-2sin^2t) or 2(cos^2t+sin^2t)... but I am struggling to remember how to differentiate these still as they all have a squared term in and I can only remember how to differentiate standard cos and sin thanks!

also, my x=sin2t, which I rearranged and differentiated to get -2costsint ... so I think best one to differentiate for y would be the one with sin and cos in... am I right in this ?
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nuodai
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Report 7 years ago
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You can differentiate 2\cos 2t straight off using the chain rule; let u=2t and so on, no need for double-angle identities (to differentiate which you also need the chain rule anyway).
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Stephhcharlene
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Report Thread starter 7 years ago
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Forgot all about chain rule... :facepalm: Thanks!
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Ree69
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NO NO NO! When differenting trig functions (that don't have a power) you just differetiate them normally but stick the derivative before it. So the derivate of 2cos2t would be 2 x 2(-sin2t) = -4sin2t. Likewise, the derivative of sin2t is 2 x cos2t = 2cos2t.

and dy/dx is simply dy/dt รท dx/dt
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Stephhcharlene
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what you just did was chain rule... o.0 lol.
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