1. ## Quotient Derivative

I'm stuck on this derivative:

y=$\displaystyle \frac{x^2 + 2ax -a}{(x+a)^2}$

Here is what I have so far:

$\displaystyle \frac{(x^2+2ax+a^2)(2x+2a) - (x^2+2ax-a)(2x+2a)}{(x+a)^4}$

$\displaystyle \frac{2a^3+4a^2-2ax}{(x+a)^4}$

$\displaystyle \frac{2a(a^2+2a-x)}{(x+a)^4}$

I know I made a mistake somewhere but I can't find it??

2. Originally Posted by dbakeg00
I'm stuck on this derivative:

y=$\displaystyle \frac{x^2 + 2ax -a}{(x+a)^2}$

Here is what I have so far:

$\displaystyle \frac{(x^2+2ax+a^2)(2x+2a) - (x^2+2ax-a)(2x+2a)}{(x+a)^4}$

$\displaystyle \frac{2a^3+4a^2-2ax}{(x+a)^4}$

$\displaystyle \frac{2a(a^2+2a-x)}{(x+a)^4}$

I know I made a mistake somewhere but I can't find it??

The expansion of

$\displaystyle (x+a)^2 is x^2+2ax+a^2$

,rather than

$\displaystyle (x^2+2ax-a)$.

3. May be that this identity allows You to save time and efforts...

$\displaystyle \frac{x^{2} + 2 a x - a}{(x+a)^{2}} = 1- \frac{a (a+1)}{(x+a)^{2}}$ (1)

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

4. Originally Posted by apcalculus
The expansion of

$\displaystyle (x+a)^2 is x^2+2ax+a^2$

,rather than

$\displaystyle (x^2+2ax-a)$.
That is how I have it on the left side of the numerator (of the first step)

$\displaystyle \frac{(x^2+2ax+a^2)(2x+2a) - (x^2+2ax-a)(2x+2a)}{(x+a)^4}$

5. Originally Posted by chisigma
May be that this identity allows You to save time and efforts...

$\displaystyle \frac{x^{2} + 2 a x - a}{(x+a)^{2}} = 1- \frac{a (a+1)}{(x+a)^{2}}$ (1)

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
Can you show me how $\displaystyle \frac{x^{2} + 2 a x - a}{(x+a)^{2}} = 1- \frac{a (a+1)}{(x+a)^{2}}$

6. Originally Posted by dbakeg00
I'm stuck on this derivative:

y=$\displaystyle \frac{x^2 + 2ax -a}{(x+a)^2}$

Here is what I have so far:

$\displaystyle \frac{(x^2+2ax+a^2)(2x+2a) - (x^2+2ax-a)(2x+2a)}{(x+a)^4}$

$\displaystyle \frac{2a^3+4a^2-2ax}{(x+a)^4}$

$\displaystyle \frac{2a(a^2+2a-x)}{(x+a)^4}$

I know I made a mistake somewhere but I can't find it??
$\displaystyle \frac{2(x+a)[(x^2+2ax+a^2) - (x^2+2ax-a)]}{(x+a)^4}=$

$\displaystyle \frac{2 [x^2-x^2+2ax-2ax+a^2+a]}{(x+a)^3}$

$\displaystyle \frac{2a(a+1)}{(x+a)^3}$

7. Originally Posted by dbakeg00
Can you show me how $\displaystyle \frac{x^{2} + 2 a x - a}{(x+a)^{2}} = 1- \frac{a (a+1)}{(x+a)^{2}}$
$\displaystyle \frac{x^{2} + 2 a x - a}{(x+a)^{2}} = \frac{x^{2} + 2 a x +a^{2} -a^{2} - a}{(x+a)^{2}} = 1 - \frac{a^{2} + a}{(x+a)^{2}}$ (1)

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

8. Originally Posted by apcalculus
$\displaystyle \frac{2(x+a)[(x^2+2ax+a^2) - (x^2+2ax-a)]}{(x+a)^4}=$

$\displaystyle \frac{2 [x^2-x^2+2ax-2ax+a^2+a]}{(x+a)^3}$

$\displaystyle \frac{2a(a+1)}{(x+a)^3}$
Thank you very much. I don't know why I didn't see that earlier. I appreciate the help!

9. Originally Posted by chisigma
$\displaystyle \frac{x^{2} + 2 a x - a}{(x+a)^{2}} = \frac{x^{2} + 2 a x +a^{2} -a^{2} - a}{(x+a)^{2}} = 1 - \frac{a^{2} + a}{(x+a)^{2}}$ (1)

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
Now I see! Is that just an identity you had memorized?