Question

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

Answer 1

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

Second derivative: `8/(x+1)^3`

Explanation:

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`

Answer 2

first derivative: `((x+3)(x-1))/(x+1)^2`
second derivative: `8/(x+1)^3`

Explanation:

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`