How doyou find the derivative of #2sinxcosx#?

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Answer 1 · 2018-04-17T15:45:57

#d/dx2sinxcosx=2cos2x#

#2sinxcosx=sin2x#

So

#d/dx2sinxcosx=d/dxsin2x=2cos2x#

Answer 2 · 2018-04-17T15:46:15

#d/dx (2sinx cosx) = 2cos 2x#

Using the linearity of the derivative and product rule this would be:

#d/dx (2sinx cosx) = 2 ( sinx (d/dx cosx) + (d/dx sinx) cosx)#

#d/dx (2sinx cosx) = 2 ( sinx (-sinx) + (cosx) cosx)#

#d/dx (2sinx cosx) = 2 (cos^2x-sin^2x)#

#d/dx (2sinx cosx) = 2 cos 2x#

but you can also use the trigonometric identity:

#2sinxcosx = sin 2x#

and then:

#d/dx (2sinx cosx) = d/dx (sin2x) = 2cos 2x#