Absolutely STUCK on a 2nd derivative

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Absolutely STUCK on a 2nd derivative!!!

1. The problem statement, all variables and given/known data

The problem is to find the second derivative of y= [tex]\sqrt{x^2+x-2}[\tex]

2. Relevant equations

I know I have to use the Chain Rule and the Quotient Rule.

3. The attempt at a solution

So, Ive gotten the first derivative (checking the book) to make sure Im 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 cant 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 Im doing with the factoring/canceling of the forementioned term?
 

Char. Limit

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Re: Absolutely STUCK on a 2nd derivative!!!

1. The problem statement, all variables and given/known data

The problem is to find the second derivative of y= [tex]\sqrt{x^2+x-2}[/tex]

2. Relevant equations

I know I have to use the Chain Rule and the Quotient Rule.

3. The attempt at a solution

So, Ive gotten the first derivative (checking the book) to make sure Im 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 cant 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 Im doing with the factoring/canceling of the forementioned term?
Fixed your latex. Remember that the right side is [/tex] not [\tex]
 
Last edited:
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Re: Absolutely STUCK on a 2nd derivative!!!

Thanks for that. Im reading on how to use LaTeX right now. Again, thanks.
 
Re: Absolutely STUCK on a 2nd derivative!!!

Hello there,

here is the problem, your second derivative is wrong it should be:

y"= [2(x^2+x-2)^(1/2)-( {((2x+1)^2)/(2(x^2+x-2)^(1/2))} )] /(2(x^2+x-2))

then what you have to do is multiply by (x^2+x-2)^(1/2) both the numerator and the denominator

thus you'll obtain this form :

y" = (2(x^2+x-2)-(2x+1)^2)/(2(x^2+x-2)^(3/2))

all you have to do now is develop the numerator of this expression and you'll get this :

y" = (-2x^2-2x-5)/(2(x^2+x-2)^(3/2))

here you go, and to get the exact one like the book, u have to multiply by 2 both numerators and denominators... Your book should have done the simplification already though...
 
307
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Re: Absolutely STUCK on a 2nd derivative!!!

Hello there,

here is the problem, your second derivative is wrong it should be:

y"= [2(x^2+x-2)^(1/2)-( {((2x+1)^2)/(2(x^2+x-2)^(1/2))} )] /(2(x^2+x-2))

then what you have to do is multiply by (x^2+x-2)^(1/2) both the numerator and the denominator

thus you'll obtain this form :

y" = (2(x^2+x-2)-(2x+1)^2)/(2(x^2+x-2)^(3/2))

all you have to do now is develop the numerator of this expression and you'll get this :

y" = (-2x^2-2x-5)/(2(x^2+x-2)^(3/2))

here you go, and to get the exact one like the book, u have to multiply by 2 both numerators and denominators... Your book should have done the simplification already though...

Thanks, but Im still confused. Why would you want to multiply the denominator and numerator by 2? I know it will make the expression match what I have in the book, but what would be the ultimate purpose (besides and answer match) to doing the multiplication?

Super appreciative for the help, btw.
 
Re: Absolutely STUCK on a 2nd derivative!!!

the purpose : absolutely nothing at all :) actually your book just didn't simplify the expression to the fullest... if you divide by 2 the numerator and deminator in the final expression reached from the book, you get to the one I found... like I said, they should have done the simplification themselves :)

and you're welcome ! anytime :)
 
307
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Re: Absolutely STUCK on a 2nd derivative!!!

Thank you so much. I've been pacing around all morning trying to figure out why that looked the way it did.

And now...on to yet more problems.

THANKS!!
 

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