How do you differentiate #y = 1/2 x + 1/4sin2x#?

1 Answer
Dec 20, 2016

#dy/dx = cos^2x#

Explanation:

This can be rewritten as follows using the identity #sin2theta = 2sinthetacostheta# and a little algebra.

#y = 1/2x + 1/4(2sinxcosx)#

#y = 1/2x + 1/2sinxcosx#

#y = 1/2(x + sinxcosx)#

#2y = x + sinxcosx#

Use implicit differentiation and the product rule to differentiate.

#2(dy/dx) = 1 + cosx(cosx) + sinx(-sinx)#

#2(dy/dx) = 1 + cos^2x - sin^2x#

Use the identity #cos^2theta + sin^2theta = 1 ->1 - sin^2theta = cos^2theta#:

#2(dy/dx) = cos^2x + cos^2x#

#2(dy/dx) = 2cos^2x#

#dy/dx = (2cos^2x)/2#

#dy/dx= cos^2x#

Hopefully this helps!