How do you find the first and second derivatives of #y= (x^2 + 2x + 5) / (x + 1)# using the quotient rule?

Question

Emilia Aguilar · Accepted Answer

First derivative: #((x+3)(x-1))/(x+1)^2#

Second derivative: #8/(x+1)^3# The rule for quotients is: #(f/g)'=(f'g-fg')/g^2# And in our problem, #f# is the numerator, and #g# is the denominator of #(x^2+2x+5)/(x+1)#. Applying the guideline: #((x^2+2x+5)/(x+1))'=((x^2+2x+5)'(x+1)-(x^2+2x+5)(x+1)')/(x+1)^2# #=((2x+2)(x+1)-(x^2+2x+5)(1))/(x+1)^2# #=(2x^2+4x+2-x^2-2x-5)/(x+1)^2# #=(x^2+2x-3)/(x+1)^2# #=((x+3)(x-1))/(x+1)^2# That is the first derivative. Use this new fraction along with the same steps to find the next one: #(((x+3)(x-1))/(x+1)^2)'=(((x+3)(x-1))'(x+1)^2-((x+3)(x-1))(x+1)^2')/((x+1)^2)^2# After extensive simplification: #8/(x+1)^3#

Scarlett Allison · Answer

first derivative: #((x+3)(x-1))/(x+1)^2#
second derivative: #8/(x+1)^3# Rule of the quotient: #(u/v)' = (u'v - v'u)/(v^2)# #u(x) = x^2 + 2x + 5# power rule: #(x^n)' = nx^(n-1)# #u(x) = x^2 + 2x + 5# #u'(x) = 2x^1 + 2x^0 = 2x + 2# #v(x) = x + 1# #v'(x) = 1x^0 = 1# #u'v = (2x + 2) * (x + 1) = 2x^2 + 4x + 2# #v'u = 1 * (x^2 + 2x + 5) = x^2 + 2x + 5# #v^2 = (x+1)^2# #(u'v - v'u) = (2x^2 + 4x + 2) - (x^2 + 2x + 5)# #= 2x^2 + 4x + 2 - x^2 - 2x - 5# #= x^2 + 2x - 3# #(u'v - v'u)/(v^2) = ((x^2+2x-3))/(x+1)^2# the first derivative of #y = (x^2+2x+5)/(x+1)# is #(x^2+2x-3)/(x+1)^2# - the quotient rule #(u/v)' = (u'v - v'u)/(v^2)# can be used again where #u(x) = x^2 + 2x -3# and #v(x) = (x+1)^2, or x^2+2x+1#. #u'(x) = 2x^1 + 2x^0 = 2x + 2, or 2(x+1)# #v'(x) = 2x^1 + 2x^0 = 2x + 2, or 2(x+1)# #u'v = (2(x+1)) * (x+1)^2 = 2(x+1)^3 = 2x^3 + 6x^2 + 6x + 2# #v'u = (2x+2) * (x^2+2x-3) = 2x^3 + 2x^2 + 4x^2 + 4x -6x - 6# #= 2x^3 + 6x^2 - 2x - 6# #u'v - v'u = (2x^3 + 6x^2 + 6x + 2) - (2x^3 + 6x^2 - 2x - 6)# #= 2x^3 + 6x^2 + 6x + 2 - 2x^3 - 6x^2 + 2x + 6# #= 8x + 8# #(u'v - v'u)/v^2 = (8x + 8)/((x+1)^4# #8x + 8 = 8(x+1)# #(8x+8)/(x+1)^4 = (8(x+1))/(x+1)^4# #= 8/(x+1)^3# the second derivative of #y = (x^2+2x+5)/(x+1)# is #8/(x+1)^3#

Luke Alvarez · Answer

undefined To find the first derivative of $ y = \frac{x^2 + 2x + 5}{x + 1} $ using the quotient rule, we differentiate the numerator and denominator separately and then apply the quotient rule formula $ \frac{d(u/v)}{dx} = \frac{v \cdot du/dx - u \cdot dv/dx}{v^2} $.

1. Differentiate the numerator:
   $ \frac{d}{dx}(x^2 + 2x + 5) = 2x + 2 $

2. Differentiate the denominator:
   $ \frac{d}{dx}(x + 1) = 1 $

3. Apply the quotient rule:
   $ \frac{d}{dx}\left(\frac{x^2 + 2x + 5}{x + 1}\right) = \frac{(x + 1)(2x + 2) - (x^2 + 2x + 5)(1)}{(x + 1)^2} $

4. Simplify the expression:
   $ \frac{(2x^2 + 2x + 2x + 2) - (x^2 + 2x + 5)}{(x + 1)^2} $
   $ = \frac{x^2 + 4x + 2 - x^2 - 2x - 5}{(x + 1)^2} $
   $ = \frac{2x - 3}{(x + 1)^2} $

Therefore, the first derivative of $ y = \frac{x^2 + 2x + 5}{x + 1} $ is $ \frac{2x - 3}{(x + 1)^2} $.