How do you differentiate #f(x)=sin^4 3x-cos^4 3x#?

1 Answer
Nov 4, 2016

Two methods are shown below.

Explanation:

Differentiate then simplify

Recall the #sin^4(3x) = (sin(3x))^4#

so the derivative is #4sin^3(3x)cos(3x)*3# (Use the chain rule twice.)

#f'(x) = 4sin^3(3x)cos(3x) * 3 - 4cos^3(3x)(-sin(3x))*3#

# = 12sin^3(3x)cos(3x) + 12cos^3(3x)sin(3x)#

# = 12sin(3x)cos(3x)underbrace((sin^2(3x) + cos^2(3x)))_(=1)#

# = 12sin(3x)cos(3x)#

Simplify then differentiate

#f(x) = sin^4(3x) - cos^4(3x)#

Factor using #a^4-b^4 = (a^2+b^2)(a^2-b^2)# (Difference of squares.)

#f(x) = sin^4(3x) - cos^4(3x)#

# = underbrace((sin^2(3x) + cos^2(3x)))_(=1)(sin^2(3x) - cos^2(3x))#

# = (sin^2(3x) - cos^2(3x))#

Now differentiate using the chain rule as in the other method

#f'(x) = 2sin(3x)cos(3x)*3-2cos(3x)(-sin(3x)*3#

# = 6sin(3x)cos(3x) + 6 sin(3x)cos(3x)#

# = 12sin(3x)cos(3x)#