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

2 Answers
Aug 14, 2015

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

Explanation:

# 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 #

Aug 15, 2015

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)#