How do you differentiate #cos 2x#?

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Answer 1 · 2017-01-03T01:13:00

We let #y = cosu# and #u = 2x#. Then #dy/(du) = -sinu# and #(du)/dx= 2#.

#dy/dx = dy/(du) xx (du)/dx#

#dy/dx = -sinu xx 2#

#dy/dx = -2sinu#

Since #u = 2x#:

#dy/dx = -2sin2x#

Hopefully this helps!