How do you find the second derivative of $y = A sin (B x)$ $y=Asin(Bx)$ ?

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Answer 1 · 2017-01-13T22:33:41

$\frac{d^{2} y}{{d x}^{2}} = - A B^{2} sin (B x)$ $(d^2y)/dx^2 = -AB^2sin(Bx)$

We use

$\frac{d}{d x} sin (K x) = K cos (K x)$ $d/dx sin(Kx) = Kcos(Kx)$ and $\frac{d}{d x} cos (K x) = - K sin (K x)$ $d/dx cos(Kx) = -Ksin(Kx)$

Differentiating:

$y = A sin (B x)$ $y = Asin(Bx)$

once gives us:

$\frac{d y}{d x} = A (B cos (B x))$ $dy/dx = A(Bcos(Bx))$
$\ \ \ \ \ = ABcos(Bx)$

Differentiating again we get:

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

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