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

Question

Ezra Adams · Accepted Answer

f'(x) =#(x-2)/x^3# or #x^-2-2x^-3#
f'' (x)= #(-2x+6)/x^4# or #-2x^-3+6x^-4# 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!!!

Evelyn Agard · Answer

undefined To find the derivatives of $ \frac{{3x^2 - x + 1}}{{x^2}} $ using the quotient rule:

1. First, identify $ u = 3x^2 - x + 1 $ and $ v = x^2 $.
2. Then, apply the quotient rule formula: $ \left( \frac{{u}}{{v}} \right)' = \frac{{u'v - uv'}}{{v^2}} $.
3. Find the derivatives of $ u $ and $ v $ with respect to $ x $.
4. Substitute the derivatives into the quotient rule formula.
5. Simplify the expression to obtain the first derivative.
6. To find the second derivative, differentiate the first derivative obtained in step 5 with respect to $ x $.