How do you use the product rule to differentiate # sin(x^2)(cos(x^2)) #?

3 Answers
May 14, 2017

Answer:

#2xcos^2(x^2)-2xsin^2(x^2)#

Explanation:

#d/dx[sin(x^2)cos(x^2)]=(sin(x^2))(d/dx[cos(x^2)])+(d/dx[sin(x^2)])(cos(x^2))= (sin(x^2))(-2xsin(x^2))+(2xcos(x^2))(cos(x^2))=2xcos^2(x^2)-2xsin^2(x^2)#

Don't forget to chain rule #x^2#

May 14, 2017

Answer:

# 2x*cos(2*x^2)#

Explanation:

Let #f(x) = sin(x^2)# and #g(x) = cos(x^2) #

You have :
# f'(x) = 2x*cos(x^2) #
and
# g'(x) = -2x*sin(x^2) #

then you apply the formula :
# f'(x)*g(x) + f(x)*g'(x) #

which gives you :
# 2x*(cos^2(x^2)-sin^2(x^2))#
Hence, with the formula #cos(2a) = cos^2 a - sin^2 a#,
# 2x*cos(2*x^2)#

May 14, 2017

The differentiation of a product is determined by applying this method:
#" "#
#d/dx (f (x)xxg (x))=d/dxf (x)xxg(x)+d/dxg (x)xxf (x)#
#" "#
#(sin (x^2)(cos (x^2)))'#
#" "#
#(sin (x^2))'xxcos (x^2)+(cosx^2)'xxsinx^2#
#" "#
#=2xcosx^2xxcosx^2-2xsinx^2xxsinx^2#
#" "#
#=2x (cosx^2xxcosx^2-sinx^2xxsinx^2)#
#" "#
#=2x (cos^2 (x^2)-sin^2 (x^2))#
#" "#
#=2xcos (2x^2)#