1. ## Quotient Derivative

I need some help on a quotient derivative:

$\frac{a+\sqrt{x}}{a-\sqrt{x}}$

Here is what I have so far:

1. $\frac{(a-x^{1/2})(\frac{1}{2}x^{-1/2}) - (a+x^{1/2})(\frac{-1}{2}x^{-1/2})}{(a-\sqrt{x})^2}$

2. $\frac{\frac{1}{2} ax^\frac{-1}{2} - \frac{1}{2}x^0 +\frac{1}{2} ax^\frac{-1}{2}- \frac{1}{2}x^0}{(a-\sqrt{x})^2}$

I know my algebra must be wrong because I'm coming up with a totally different answer than the book. Could someone please help give me a play by play?

Thanks
-db

2. Originally Posted by dbakeg00
I need some help on a quotient derivative:

$\frac{a+\sqrt{x}}{a-\sqrt{x}}$

Here is what I have so far:

1. $\frac{(a-x^{1/2})(\frac{1}{2}x^{-1/2}) - (a+x^{1/2})(\frac{-1}{2}x^{-1/2})}{(a-\sqrt{x})^2}$

2. $\frac{\frac{1}{2} ax^\frac{-1}{2} - \frac{1}{2}x^0 +\frac{1}{2} ax^\frac{-1}{2}- \frac{1}{2}x^0}{(a-\sqrt{x})^2}$

I know my algebra must be wrong because I'm coming up with a totally different answer than the book. Could someone please help give me a play by play?

Thanks
-db
Multiply by $\frac{a+\sqrt{x}}{a+ \sqrt{x}}$

$\frac{a+\sqrt{x}}{a-\sqrt{x}} \times \frac{a+\sqrt{x}}{a+ \sqrt{x}} = \frac{(a+\sqrt{x})^2}{a^2-x}$

$u = (a + \sqrt{x})^2 \: \rightarrow \: u' = a+\sqrt{x}$

$v = a^2-x \: \rightarrow \: v' = -1$

$y' = \frac{u'v-v'u}{v^2} = \frac{(a^2-x)(a+\sqrt{x}) + (a+\sqrt{x})^2}{(a^2-x)^2} = \frac{(a+\sqrt{x})(a^2-x+a+\sqrt{x})}{(a^2-x)^2}$

And that's the form I'd leave it in.

3. Thank you. 2 questions though:

1. How did you know to multiply by $

\frac{a+\sqrt{x}}{a+ \sqrt{x}}
$

2. the book is giving this answer:
$\frac{a}{(a-\sqrt{x})^2*\sqrt{x}}$

4. Originally Posted by dbakeg00
Thank you. 2 questions though:

1. How did you know to multiply by $

\frac{a+\sqrt{x}}{a+ \sqrt{x}}
$

2. the book is giving this answer:
$\frac{a}{(a-\sqrt{x})^2*\sqrt{x}}$
I didn't, that was just a guess based on rationalising the denominator. Turned out to be wrong.

Note that $x^{-1/2} = \frac{1}{\sqrt{x}}$

$u = a+\sqrt{x} \: \: \rightarrow \:\: u' = \frac{1}{2\sqrt{x}}$

$v = a - \sqrt{x} \: \: \rightarrow \: \: v' = -\frac{1}{2\sqrt{x}}$

$y' = \frac{\frac{1}{2\sqrt{x}}(a-\sqrt{x}) + \frac{1}{2\sqrt{x}}(a+\sqrt{x})}{(a - \sqrt{x})^2}$

Note that in the numerator there is a factor of $\frac{1}{\sqrt{x}}$

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

Because $\frac{a}{b} \times \frac{c}{d} = \frac{ac}{bd}$ we can put that $2\sqrt{x}$ on the denominator. Also $\sqrt{x}$ cancels in the numerator

$\frac{2a}{2\sqrt{x} \times (a-\sqrt{x})^2}$

Lastly 2 cancels to give the book's answer of $\frac{a}{\sqrt{x} \cdot (a-\sqrt{x})^2}$

5. Originally Posted by dbakeg00
Thank you. 2 questions though:

1. How did you know to multiply by $

\frac{a+\sqrt{x}}{a+ \sqrt{x}}
$

2. the book is giving this answer:
$\frac{a}{(a-\sqrt{x})^2*\sqrt{x}}$

$
\frac{(a-x^{1/2})(\frac{1}{2}x^{-1/2}) - (a+x^{1/2})(\frac{-1}{2}x^{-1/2})}{(a-\sqrt{x})^2}=\frac{\frac{1}{2} ax^\frac{-1}{2} - \frac{1}{2}x^0 +\frac{1}{2} ax^\frac{-1}{2}+ \frac{1}{2}x^0}{(a-\sqrt{x})^2}$

$=\frac{ax^\frac{-1}{2}}{(a-\sqrt{x})^2}$

$=\frac{a}{(a-\sqrt{x})^2*\sqrt{x}}$

6. Originally Posted by ione
If you simplify you original answer, you will get the answer given in the book

$
\frac{\frac{1}{2} ax^\frac{-1}{2} - \frac{1}{2}x^0 +\frac{1}{2} ax^\frac{-1}{2}- \frac{1}{2}x^0}{(a-\sqrt{x})^2}=\frac{ax^\frac{-1}{2}}{(a-\sqrt{x})^2}$

$=\frac{a}{(a-\sqrt{x})^2*\sqrt{x}}$
In the numerator, I see where you are getting the "-1" but I don't understand where the "+1" is coming from??

7. Originally Posted by dbakeg00
In the numerator, I see where you are getting the "-1" but I don't understand where the "+1" is coming from??