How do you differentiate #sinxcosx#?
1 Answer
Mar 9, 2017
# d/dx sinxcosx = cos2x#
Explanation:
Method 1 - Manipulation
# d/dx sinxcosx = d/dx 1/2 sin 2x #
# " " = 1/2 d/dx sin 2x #
# " " = 1/2 2cos2x #
# " " = cos2x #
Method 2: Product Rule
# d/dx sinxcosx = (sinx)(d/dxcosx)+(d/dxsinx)(cosx) #
# " " = (sinx)(-sinx)+(cosx)(cosx)#
# " " = cos^2x-sin^2x#
# " " = cos2x#
