Second derivative of (x^2-4)/(x+1)

[TayTayDatDude]

Homework Statement

Determine the second derivative for the following function:

y= (x^2-4)/(x+1)

Homework Equations

The Attempt at a Solution

Well, the first derivative is (x^2+2x+4)/(x^2+2x+1)

For calculating the second derivative I can only get as far as (-6x-6)/(x^4+4x^3+6x^2+4x+1)

On an online derivative calculator, the answer is : -(6/(x^3+3x^2+3x+1)) , yet I can't get to it :(

 

[quantumdude]

The Attempt at a Solution

Well, the first derivative is (x^2+2x+4)/(x^2+2x+1)

It would be helpful for you to factor the denominator, and then try to take the second derivative.

For calculating the second derivative I can only get as far as (-6x-6)/(x^4+4x^3+6x^2+4x+1)

On an online derivative calculator, the answer is : -(6/(x^3+3x^2+3x+1)) , yet I can't get to it :(

Your answer is correct. However you need to reduce the fraction to get the result from the online calculator. This would be easiest if you follow my suggestion above: factor the denominator in f'.

 

[Mark44]

Well, the first derivative is (x^2+2x+4)/(x^2+2x+1)

So $$dy/dx = \frac{x^2 + 2x + 1 + 3}{(x + 1)^2} = \frac{(x + 1)^2 + 3}{(x + 1)^2}$$
$$= 1 + \frac{3}{(x + 1)^2} = 1 + 3(x + 1)^{-2}$$

The extra work it took to get the derivative into this form is more than made up by the time saved in getting the next derivative, which can be done by a fairly simple application of the chain rule.

 

[TayTayDatDude]

So $$dy/dx = \frac{x^2 + 2x + 1 + 3}{(x + 1)^2} = \frac{(x + 1)^2 + 3}{(x + 1)^2}$$
$$= 1 + \frac{3}{(x + 1)^2} = 1 + 3(x + 1)^{-2}$$

The extra work it took to get the derivative into this form is more than made up by the time saved in getting the next derivative, which can be done by a fairly simple application of the chain rule.

I have not yet learned the chain rule, and I do not know how to find the derivative of (x+1)^2, other than by expanding it.

so, (-6x-6)/(x+1)^4

 

[TayTayDatDude]

Edit, solved, thanks.