Question

Help solving this derivatives problem please?

Answer 1

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:

$\dfrac{d}{dx}(-\sin{(2x)})=-2\cos{(2x)}\\\\\dfrac{d^2}{dx^2}(-\sin{(2x)})=2^2\sin{(2x)}\\\\\dfrac{d^3}{dx^3}(-\sin{(2x)})=2^3\cos{(2x)}\\\\\dfrac{d^4}{dx^4}(-\sin{(2x)})=-2^4\sin{(2x)}$

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)