Question

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

Answer 1

f'(x) =`(x-2)/x^3` or `x^-2-2x^-3`
f'' (x)= `(-2x+6)/x^4` or `-2x^-3+6x^-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
`(v/u)'= (vu'-uv')/v^2`

For the question u=`3x^2-x +1`
u'= `6x-1`
v=`x^2`
v'=2`x`
This the first derivative of our quotient is
`{x^2(6x-1)-(3x^2-x+1)2x}/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!!!