How doyou find the derivative of #2sinxcosx#?

2 Answers
Apr 17, 2018

Answer:

#d/dx2sinxcosx=2cos2x#

Explanation:

#2sinxcosx=sin2x#

So

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

Apr 17, 2018

Answer:

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

Explanation:

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#