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

Question

Emma Aldridge · Accepted Answer

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

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

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

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

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

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

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

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

We can readjust the three terms:

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

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

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

The result will be the sum of the three terms.

Cameron Alexander · Answer

undefined To differentiate $ g(x) = 3\tan(4x) \cdot \sin(2x) \cdot \cos(2x) $ using the product rule:

1. Identify the functions $ f(x) = 3\tan(4x) $, $ h(x) = \sin(2x) $, and $ k(x) = \cos(2x) $.
2. Apply the product rule: $ g'(x) = f'(x) \cdot h(x) \cdot k(x) + f(x) \cdot h'(x) \cdot k(x) + f(x) \cdot h(x) \cdot k'(x) $.
3. Find the derivatives of each function using standard differentiation rules.
4. Substitute the derivatives and original functions into the product rule formula to find $ g'(x) $.