How do you differentiate #g(x)= 3tan4x *sin2x*cos2x# using the product rule?

1 Answer
Nov 6, 2015

Answer:

# (12sin(2x)cos(2x))/(cos^2(4x))+6tan(4x)*cos^2(2x)-6tan(4x)sin^2(2x)#

Explanation:

The rule is very simple, you derive one function and leave the others untouched, and then sum all these terms. So:

  • Derive the first function and leave the other two untouched:

#(d/dx 3tan(4x))*sin(2x)*cos(2x)#

#= (3 * 1/(cos^2(4x))*4)*sin(2x)*cos(2x)#

  • Derive the second function and leave the other two untouched:

# 3tan(4x)* d/dx sin(2x)*cos(2x)#

#= 3tan(4x)* (cos(2x)* 2)* cos(2x)#

  • Derive the third function and leave the other two untouched:

# 3tan(4x)* sin(2x)* d/dx cos(2x)#

#= 3tan(4x)* sin(2x)* (-sin(2x)*2)#

We can readjust the three terms:

  1. #(3 * 1/(cos^2(4x))*4)*sin(2x)*cos(2x) = (12sin(2x)cos(2x))/(cos^2(4x))#

  2. #3tan(4x)* (cos(2x)* 2)* cos(2x) = 6tan(4x)*cos^2(2x)#

  3. #3tan(4x)* sin(2x)* (-sin(2x)*2) = -6tan(4x)sin^2(2x)#

The result will be the sum of the three terms.