Question

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

Answer 1

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)`