Calc BC derivative problem with trig and double angle -- Help please

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To find the derivative f'(x) of the function f(x) = 8^(sin^2(3x)), the double angle formula for sine is necessary. The derivative is calculated using the formula y' = ln a * a^u * du, resulting in f'(x) = ln 8 * 8^(sin^2(3x)) * 3sin(6x). The challenge arises in simplifying the expression to contain only one trigonometric function, which may involve using the identity sin(2θ) = 2sin(θ)cos(θ). The discussion highlights uncertainty about the requirement for simplification and the application of the double angle formula. Understanding how to express the derivative with a single trig function is key to solving the problem correctly.
jessieb128
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Homework Statement


Find f'(x) if f(x) = 8^(sin^2(3x))
Hint: you will need to use the double angle formula for trig functions and your answer should only have one trig function in it.

Homework Equations


if y=a^u then y' = ln a * a^u * du
sin(2x) = 2sinxcosx

The Attempt at a Solution


We're only on the second chapter so I'm not sure if we really need the first equation I gave, which is from the fifth chapter (but I'm in BC so maybe she expects us to know it already?). Still, using that equation I got f'(x) = ln 8 * 8^(sin^2(3x)) * 3sin6x

The 3sin6x came from when I took the derivative of sin^2(3x) and converted it from 6sin3xcos3x to 3sin6x

But I don't know how my teacher (an online teacher) wanted the answer to only have one trig function? Did I mess up somewhere?
 
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In terms of sin2θ, cos(2θ)=??
 
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