Answer:
-2^80·sin(2x)
Step-by-step explanation:
First of all, the problem can be simplified a bit by recognizing the symmetry of the sine function:
sin(-2x) = -sin(2x)
Then, you need to recognize the "periodic" nature of the derivatives of the sine function:
[tex]\dfrac{d}{dx}(-\sin{(2x)})=-2\cos{(2x)}\\\\\dfrac{d^2}{dx^2}(-\sin{(2x)})=2^2\sin{(2x)}\\\\\dfrac{d^3}{dx^3}(-\sin{(2x)})=2^3\cos{(2x)}\\\\\dfrac{d^4}{dx^4}(-\sin{(2x)})=-2^4\sin{(2x)}[/tex]
That is, the coefficient of x gets raised to the power of the derivative number, and the function (sin, cos) cycles with a repeat of 4. Thus, the 80th derivative is ...
-2^80·sin(2x)
