How do you find the derivative of #y=(x^2+1)^-1(3x-1)^-2#?

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Answer 1 · 2017-03-24T09:55:50

#y' =(-12x^2 +2x - 6)/((x^2 + 1)^2(3 x - 1)^(3))#

#y = (x^2 + 1)^(-1) (3 x - 1)^(-2)#

Let say #u =(x^2 + 1)^(-1), u' = (-1)*(x^2 + 1)^(-2)*2x=-2x(x^2 + 1)^(-2)#
and,

#v= (3 x - 1)^(-2), v' = (-2)*(3 x - 1)^(-3)*3 = -6(3 x - 1)^(-3)#

Therefore,
#y = u^-1 v^-2#

#y' = uv' + u'v#

#y' = (x^2 + 1)^(-1)*-6(3 x - 1)^(-3) + (3 x - 1)^(-2)-2x(x^2 + 1)^(-2)#

#y' = -6(x^2 + 1)^(-1)(3 x - 1)^(-3) -2 x(3 x - 1)^(-2)(x^2 + 1)^(-2)#

#y' =(-6)/((x^2 + 1)(3 x - 1)^(3)) -(2 x)/((3 x - 1)^(2)(x^2 + 1)^(2))#

#y' =((-6)(x^2 + 1))/((x^2 + 1)^2(3 x - 1)^(3)) -((2 x)(3x -1))/((3 x - 1)^(3)(x^2 + 1)^(2))#

#y' =((-6x^2 - 6))/((x^2 + 1)^2(3 x - 1)^(3)) -((6x^2 - 2x))/((3 x - 1)^(3)(x^2 + 1)^(2))#

#y' =(-6x^2 - 6 -6x^2 +2x)/((x^2 + 1)^2(3 x - 1)^(3))#

#y' =(-12x^2 +2x - 6)/((x^2 + 1)^2(3 x - 1)^(3))#