How do you find the second derivative of #y=Asin(Bx)#?

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Answer 1 · 2017-01-13T22:33:41

#(d^2y)/dx^2 = -AB^2sin(Bx)#

We use

#d/dx sin(Kx) = Kcos(Kx)# and #d/dx cos(Kx) = -Ksin(Kx)#

Differentiating:

#y = Asin(Bx)#

once gives us:

#dy/dx = A(Bcos(Bx))#
#\ \ \ \ \ = ABcos(Bx)#

Differentiating again we get:

#(d^2y)/dx^2 = AB(-Bsin(Bx))#
# \ \ \ \ \ \ \ = -AB^2sin(Bx)#