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

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Answer 1 · 2016-11-04T01:57:26

Two methods are shown below.

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

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

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

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

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