# How do you find the first and second derivatives of ((3x^2-x+1)/(x^2)) using the quotient rule?

Aug 22, 2016

f'(x) =$\frac{x - 2}{x} ^ 3$ or ${x}^{-} 2 - 2 {x}^{-} 3$
f'' (x)= $\frac{- 2 x + 6}{x} ^ 4$ or $- 2 {x}^{-} 3 + 6 {x}^{-} 4$

#### Explanation:

u is a function of X, u' is the first derivative
v is a function of x, v' is the first derivative
The quotient rule gives
$\left(\frac{v}{u}\right) ' = \frac{v u ' - u v '}{v} ^ 2$

For the question u=$3 {x}^{2} - x + 1$
u'= $6 x - 1$
v=${x}^{2}$
v'=2$x$
This the first derivative of our quotient is
$\frac{{x}^{2} \left(6 x - 1\right) - \left(3 {x}^{2} - x + 1\right) 2 x}{x} ^ 4$
And now tidy up to the answer
BUT why?
Write the function as 3-${x}^{-} 1 + {x}^{-} 2$
And differentiate each term.
Mathematicians are lazy..do things the most efficient way!!!