What is the derivative of #sin(2x)#?

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Answer 1 · 2015-01-25T11:43:57

#2*cos(2x)#

I would use the Chain Rule:

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First derive #sin# and then the argument #2x# to get:
#cos(2x)*2#

Answer 2 · 2018-06-20T23:28:30

#2cos2x#

The key realization is that we have a composite function, which can be differentiated with the help of the Chain Rule

#f'(g(x))*g'(x)#

We essentially have a composite function

#f(g(x))# where

#f(x)=sinx=>f'(x)=cosx# and #g(x)=2x=>g'(x)=2#

We know all of the values we need to plug in, so let's do that. We get

#cos(2x)*2#

#=>2cos2x#

Hope this helps!