# Differentiating 2cos2t (Parametic Equations)Watch

#1
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 ?
0
7 years ago
#2
You can differentiate straight off using the chain rule; let and so on, no need for double-angle identities (to differentiate which you also need the chain rule anyway).
#3
Forgot all about chain rule... Thanks!
0
7 years ago
#4
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 Ã· dx/dt
0
#5
what you just did was chain rule... o.0 lol.
0
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