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

The quotient rule 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)#.

Using the rule:

#((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's the first derivative. To find the next one, perform the same actions, but with this new fraction:

#(((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 a lot of simplifying:

#8/(x+1)^3#

quotient rule:

#(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#