What is the second derivative of the function #f(x) = (x) / (x - 1)#?

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Answer 1 · 2016-10-06T03:34:18

#d^2/(dx^2)x/(x-1)=2/(x-1)^3#

For this problem, we will use the quotient rule:

#d/dx f(x)/g(x) = (g(x)f'(x)-f(x)g'(x))/[g(x)]^2#

We can also make it a little easier by dividing to get

#x/(x-1) = 1+1/(x-1)#

First derivative:

#d/dx(1+1/(x-1))#

#= (d/dx1)+(d/dx((x-1)(d/dx1)-1(d/dx(x-1)))/(x-1)^2)#

#=0+((x-1)(0)-(1)(1))/(x-1)^2#

#= -1/(x-1)^2#

Second derivative:

The second derivative is the derivative of the first derivative.

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

#=-((x-1)^2(d/dx1)-1(d/dx(x-1)^2))/[(x-1)^2]^2#

#=-((x-1)^2(0)-1(2(x-1)))/(x-1)^4#

#=2/(x-1)^3#

We could also have used the power rule #d/dx x^n = nx^(n-1)# for #n!=1#:

#1+1/(x-1) = 1+(x-1)^(-1)#

#=> d/dx (1+1/(x-1)) =d/dx(1+(x-1)^(-1))#

#= -(x-2)^(-2)#

#=> d^2/(dx^2)(1+1/(x-1)) = d/dx(-(x-2)^(-2))#

#=2(x-2)^(-3)#

which is the same as the result we obtained above.