How do you find the second derivative of $sin(x)/(1-cos(x))$ ?

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Answer 1 · 2015-08-14T13:22:57

$d^2/dx^2(sin x/(1-cos x)) = sin x/(cos x - 1)^2$

$d/dx(sin x/(1-cos x)) = ((1-cos x)cos x - sin^2 x)/(1-cos x)^2 = 1/(cos x-1)$
$d^2/dx^2(sin x/(1-cos x)) = sin x/(cos x - 1)^2$

Answer 2 · 2015-08-15T21:52:58

You take the first derivative twice.

$color(green)(d^2/(dx^2)[f(x)] = d/(dx)[d/(dx)[f(x)]])$

$= d/(dx)[((1-cosx)(cosx) - sinx(0+sinx))/(1-cosx)^2]$

$= d/(dx)[(cosx-cos^2x - sin^2x)/(1-cosx)^2]$

$= d/(dx)[(cosx-(cos^2x + sin^2x))/(1-cosx)^2]$

$= d/(dx)[(-cancel((1-cosx)))/(1-cosx)^cancel(2)]$

$= d/(dx)[-1/(1-cosx)] = d/(dx)[-(1-cosx)^-1]$

$= (1-cosx)^(-2)*sinx$

$= color(blue)(sinx/(1-cosx)^2)$

How do you find the second derivative of sin(x)/(1-cos(x))sin(x)/(1-cos(x))?