Absolutely STUCK on a 2nd derivative

[mateomy]

Absolutely STUCK on a 2nd derivative!

Homework Statement

The problem is to find the second derivative of y= \sqrt{x^2+x-2}[\tex]<br /> <br /> <h2>Homework Equations</h2><br /> <br /> I know I have to use the Chain Rule and the Quotient Rule.<br /> <br /> <h2>The Attempt at a Solution</h2><br /> <br /> So, I&#039;ve gotten the first derivative (checking the book) to make sure I am right...<br /> <br /> y&#039;= \frac{\2x+1}{2\sqrt{x^2+x-2}[\tex]&lt;br /&gt; &lt;br /&gt; Moving on...I&amp;#039;ve managed to set up the 2nd as so...&lt;br /&gt; &lt;br /&gt; y&amp;#039;&amp;#039;=\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]&amp;lt;br /&amp;gt; &amp;lt;br /&amp;gt; And this is where I get stuck, because the book ends up getting something like...&amp;lt;br /&amp;gt; &amp;lt;br /&amp;gt; y&amp;amp;#039;&amp;amp;#039;=\frac{-4x^2-4x-10}{4(x^2+x-2)^3/2}[\tex]&amp;amp;lt;br /&amp;amp;gt; &amp;amp;lt;br /&amp;amp;gt; I can&amp;amp;amp;#039;t figure out how they get that in the end. When I factor out the (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?

 

[Char. Limit]

Homework Statement

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

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'= \frac{2x+1}{2\sqrt{x^2+x-2}}

Moving on...I've managed to set up the 2nd as so...

y''=\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}

And this is where I get stuck, because the book ends up getting something like...

y''=\frac{-4x^2-4x-10}{4(x^2+x-2)^3/2}

I can't figure out how they get that in the end. When I factor out the (x^2+x-2) from the top and bottom my numerator looks waaaay off. Is it something I am doing with the factoring/canceling of the forementioned term?

Fixed your latex. Remember that the right side is [/tex] not [\tex]

 

[mateomy]

Thanks for that. I am reading on how to use LaTeX right now. Again, thanks.

 

[joeghal87]

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...

 

[mateomy]

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 I am 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.

 

[joeghal87]

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 :)

 

[mateomy]

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!