How do you find the derivative of #sin2x - cos2x#?
1 Answer
Mar 12, 2016
2cos2x + 2sin2x
Explanation:
differentiate using
#color(blue)" chain rule " #
#d/dx [ f(g(x)) ] = f'(g(x)). g'(x)# apply this rule to each term
#d/dx(sin2x) = cos2x d/dx(2x) = 2cos2x #
#d/dx(cos2x) = -sin2x d/dx(2x) = -2sin2x #
#rArrd/dx(sin2x-cos2x) = 2cos2x + 2sin2x#
