- #1
mateomy
- 307
- 0
Absolutely STUCK on a 2nd derivative!
The problem is to find the second derivative of y= [tex]\sqrt{x^2+x-2}[\tex]
I know I have to use the Chain Rule and the Quotient Rule.
So, I've gotten the first derivative (checking the book) to make sure I am right...
y'= [tex]\frac{\2x+1}{2\sqrt{x^2+x-2}[\tex]
Moving on...I've managed to set up the 2nd as so...
y''=[tex]\frac{(x^2+x-2)^2(2)-(2x+1)(1/2)(x^2+x-2)^-1/2(2x+1)}{2(x^2+x-2)^3/2}[\tex]
And this is where I get stuck, because the book ends up getting something like...
y''=[tex]\frac{-4x^2-4x-10}{4(x^2+x-2)^3/2}[\tex]
I can't figure out how they get that in the end. When I factor out the [tex](x^2+x-2)[\tex] from the top and bottom my numerator looks waaaay off. Is it something I am doing with the factoring/canceling of the forementioned term?
Homework Statement
The problem is to find the second derivative of y= [tex]\sqrt{x^2+x-2}[\tex]
Homework Equations
I know I have to use the Chain Rule and the Quotient Rule.
The Attempt at a Solution
So, I've gotten the first derivative (checking the book) to make sure I am right...
y'= [tex]\frac{\2x+1}{2\sqrt{x^2+x-2}[\tex]
Moving on...I've managed to set up the 2nd as so...
y''=[tex]\frac{(x^2+x-2)^2(2)-(2x+1)(1/2)(x^2+x-2)^-1/2(2x+1)}{2(x^2+x-2)^3/2}[\tex]
And this is where I get stuck, because the book ends up getting something like...
y''=[tex]\frac{-4x^2-4x-10}{4(x^2+x-2)^3/2}[\tex]
I can't figure out how they get that in the end. When I factor out the [tex](x^2+x-2)[\tex] from the top and bottom my numerator looks waaaay off. Is it something I am doing with the factoring/canceling of the forementioned term?