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

Mar 24, 2017

$y ' = \frac{- 12 {x}^{2} + 2 x - 6}{{\left({x}^{2} + 1\right)}^{2} {\left(3 x - 1\right)}^{3}}$

#### Explanation:

$y = {\left({x}^{2} + 1\right)}^{- 1} {\left(3 x - 1\right)}^{- 2}$

Let say $u = {\left({x}^{2} + 1\right)}^{- 1} , u ' = \left(- 1\right) \cdot {\left({x}^{2} + 1\right)}^{- 2} \cdot 2 x = - 2 x {\left({x}^{2} + 1\right)}^{- 2}$
and,

$v = {\left(3 x - 1\right)}^{- 2} , v ' = \left(- 2\right) \cdot {\left(3 x - 1\right)}^{- 3} \cdot 3 = - 6 {\left(3 x - 1\right)}^{- 3}$

Therefore,
$y = {u}^{-} 1 {v}^{-} 2$

$y ' = u v ' + u ' v$

$y ' = {\left({x}^{2} + 1\right)}^{- 1} \cdot - 6 {\left(3 x - 1\right)}^{- 3} + {\left(3 x - 1\right)}^{- 2} - 2 x {\left({x}^{2} + 1\right)}^{- 2}$

$y ' = - 6 {\left({x}^{2} + 1\right)}^{- 1} {\left(3 x - 1\right)}^{- 3} - 2 x {\left(3 x - 1\right)}^{- 2} {\left({x}^{2} + 1\right)}^{- 2}$

$y ' = \frac{- 6}{\left({x}^{2} + 1\right) {\left(3 x - 1\right)}^{3}} - \frac{2 x}{{\left(3 x - 1\right)}^{2} {\left({x}^{2} + 1\right)}^{2}}$

$y ' = \frac{\left(- 6\right) \left({x}^{2} + 1\right)}{{\left({x}^{2} + 1\right)}^{2} {\left(3 x - 1\right)}^{3}} - \frac{\left(2 x\right) \left(3 x - 1\right)}{{\left(3 x - 1\right)}^{3} {\left({x}^{2} + 1\right)}^{2}}$

$y ' = \frac{\left(- 6 {x}^{2} - 6\right)}{{\left({x}^{2} + 1\right)}^{2} {\left(3 x - 1\right)}^{3}} - \frac{\left(6 {x}^{2} - 2 x\right)}{{\left(3 x - 1\right)}^{3} {\left({x}^{2} + 1\right)}^{2}}$

$y ' = \frac{- 6 {x}^{2} - 6 - 6 {x}^{2} + 2 x}{{\left({x}^{2} + 1\right)}^{2} {\left(3 x - 1\right)}^{3}}$

$y ' = \frac{- 12 {x}^{2} + 2 x - 6}{{\left({x}^{2} + 1\right)}^{2} {\left(3 x - 1\right)}^{3}}$