Question

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

Answer 1

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

Explanation:

`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))`