How do you use the Product Rule to find the derivative of #f(z) = z^3sin^2(2z)#?

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Answer 1 · 2015-10-19T16:54:36

See the explanation.

I assume that he challenge is to differentiate #sin^2(2z)#

Note that #sin^2(2z) = [sin(2z)]^2#

So the derivative is
#2[sin(2z)] d/dz(sin(2z)) = 2[sin(2z)] (cos(2z) d/dz(2z))#

# = 2[sin(2z)] (cos(2z) (2)#

# = 4sin(2z)cos(2z)#

This last expression can be rewritten as #2sin(4x)#

Now to the big question:

#f'(z) = 3z^2sin^2(2z) + z^3 [2sin(2z)cos(2z)*2]#

# = 3z^2sin^2(2z) + 4z^3sin(2z)cos(2z)#

Rewrite/simplify to taste.