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How do you differentiate #f(x)=2sinxcosx#?

1 Answer

Nov 2, 2016

#f'(x) = 2cos(2x)#

Explanation:

Use the identity #2sinthetacostheta = sin2theta# to rewrite the function.

#f(x) = sin2x#

Now, use the chain rule to differentiate.

Let #y = sinu# and #u = 2x#. #y' = cosu# and #u' = 2#.

#dy/dx = cosu xx 2 = cos(2x) xx 2 = 2cos(2x)#

Hopefully this helps!

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