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)#