Question

高数，求解这道题

Answer 1

y=sin2x
y'=2cos2x=2sin(2x+π/2)
y"=-4sin2x=4sin(2x+π)
y"'=-8cos2x=8sin(2x+3π/2)
y""=16sin2x=16sin(2x+2π)
.....
y^(n)=2^nsin(2x+nπ/2)

Answer 2

f(x)=sinx =>f(0) =0
f'(x)=cosx =>f'(0)/1! =1
f''(x)=-sinx =>f''(0)/2! =0
f'''(x)=-cosx =>f'''(0)/3! = -1/6
f''''(x)=sinx=f(x) =>f''''(0)/4! = 0
sinx = x - (1/3!)x^3 + (1/5!)x^5 + .....+ [(-1)^(n+1) /(2n-1)!] x^(2n-1) +....
sin2x =(2x) - (1/3!)(2x)^3 + (1/5!)(2x)^5 + .....+ [(-1)^(n+1) /(2n-1)!] (2x)^(2n-1) +....