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

1 Answer
Aug 22, 2016

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'=2x
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!!!