How do you differentiate #f(x)= sin2xtan4x# using the product rule?

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Answer 1 · 2015-12-29T08:28:04

#f'(x)=2cos2xtan4x+4sin2xsec^2 4x#

The product rule states that:

#f'(x)=tan4xd/dx(sin2x)+sin2xd/dx(tan4x)#

Finding either derivative requires the chain rule.

According to the chain rule,

#d/dx(sinu)=u'cosu# and #d/dx(tanu)=u'sec^2u#.

Thus,

#d/dx(sin2x)=d/dx(2x)cos2x=2cos2x#

#d/dx(tan4x)=d/dx(4x)sec^2 4x=4sec^2 4x#

Plug these back in.

#f'(x)=2cos2xtan4x+4sin2xsec^2 4x#