Higher Order Derivatives — Question 9

Question 9 - Solution

Successive differentiation yields

$$f'=-2\sin2x,\quad f''=-4\cos2x,\quad f^{(3)}=8\sin2x,\quad f^{(4)}=16\cos2x.$$

Each derivative multiplies by $2$ and advances the phase by $\pi/2$; induction therefore gives

$$\boxed{f^{(n)}(x)=2^n\cos(2x+n\pi/2)}.$$

Since $15\pi/2$ is congruent to $3\pi/2$ modulo $2\pi$,

$$\boxed{f^{(15)}(x)=32768\cos(2x+3\pi/2)=32768\sin2x}.$$