Question #45c8c

Question

480 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-05-15T01:07:39

#-2cos(-2x)#

I will assume you meant #sin(-2x)#:

Use chain rule. First, take the derivative of #sin(u)# where #u=-2x#. But we'll plug that in later.:

#(sin(u))'=cos(u)#

Now, we have to find #u'#.

#u'=-(-2x)'=-2#:

Multiply the two together:

#-2cos(u)#

Now plug #u# back in:

#-2cos(-2x)#