Can someone show me how to find this derivative?

**KarlosK** · April 27th 2010, 08:07 AM

I need to take the derivative of... (-x^2 - 2x) / (x^2 + 2x + 2)^2

I know I use the quotient rule which is

(x^2 + 2 x + 2)^2 (-2x - 2) - (-x^2 - 2x) 2(x^2 + 2)
------------------------------------------------------
(x^2 + 2x + 2)^4

I am having trouble simplifying this and getting to the correct answer. Can someone please show me how to do this!

**Failure** · April 27th 2010, 08:26 AM

Originally Posted by KarlosK

I need to take the derivative of... (-x^2 - 2x) / (x^2 + 2x + 2)^2

I know I use the quotient rule which is

(x^2 + 2 x + 2)^2 (-2x - 2) - (-x^2 - 2x) 2(x^2 + 2)
------------------------------------------------------
(x^2 + 2x + 2)^4

I am having trouble simplifying this and getting to the correct answer. Can someone please show me how to do this!

Your derivative is not quite correct, you need to apply the chain rule to determine the derivative of the denominator of the original fraction, then you get

$\frac{(x^2 + 2 x + 2)^2 (-2x - 2) - (-x^2 - 2x){\color{blue}2(x^2 + 2x+2)\cdot (2x+2)}}{(x^2 + 2x + 2)^4}$

And the next thing you do is cancel the two common factors $(x^2+2x+2)$ and $(2x+2)$ of numerator and denominator.

**KarlosK** · April 27th 2010, 08:41 AM

Originally Posted by Failure

Your derivative is not quite correct, you need to apply the chain rule to determine the derivative of the denominator of the original fraction, then you get

$\frac{(x^2 + 2 x + 2)^2 (-2x - 2) - (-x^2 - 2x){\color{blue}2(x^2 + 2x+2)\cdot (2x+2)}}{(x^2 + 2x + 2)^4}$

And the next thing you do is cancel the two common factors $(x^2+2x+2)$ and $(2x+2)$ of numerator and denominator.

I came up with a final answer of

2(x+1) (x^2 + 2x -2)
----------------------
(x^2 + 2x + 2)^3

**Archie Meade** · April 27th 2010, 09:55 AM

Originally Posted by KarlosK

I came up with a final answer of

2(x+1) (x^2 + 2x -2)
----------------------
(x^2 + 2x + 2)^3

That's it,

You could also try

$\frac{d}{dx}\left(\frac{-x^2-2x}{\left(x^2+2x+2\right)^2}\right)$ $=-\frac{d}{dx}\left(\frac{\left(x^2+2x+2\right)-2}{\left(x^2+2x+2\right)^2}\right)$

$=-\frac{d}{dx}\left(\frac{1}{x^2+2x+2}\right)+\frac{ d}{dx}\left(\frac{2}{\left(x^2+2x+2\right)^2}\righ t)$ $=-\frac{d}{dx}\left(x^2+2x+2\right)^{-1}+2\frac{d}{dx}\left(x^2+2x+2\right)^{-2}$

$=\left(x^2+2x+2\right)^{-2}(2x+2)-4\left(x^2+2x+2\right)^{-3}(2x+2)$

using the chain rule, giving

$\frac{2x+2}{\left(x^2+2x+2\right)^2}-\frac{4(2x+2)}{\left(x^2+2x+2\right)^3}$

$=\frac{(2x+2)\left(x^2+2x+2\right)-4(2x+2)}{\left(x^2+2x+2\right)^3}$

$=\frac{2(x+1)\left(x^2+2x+2-4\right)}{\left(x^2+2x+2\right)^2}$

$=\frac{2(x+1)\left(x^2+2x-2\right)}{\left(x^2+2x+2\right)^3}$

Or

$u=x^2+2x+2$

$u-2=x^2+2x$

then differentiate

$-\frac{d}{dx}\left(\frac{u-2}{u^2}\right)=-\frac{du}{dx}\frac{d}{du}\left(\frac{u-2}{u^2}\right)$

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