1. ## differenciation

1)

$
(x + \frac {1}{x})' = \frac {1}{2}x^{-0.5} - \frac {x^{0.5}}{x}= \frac {1}{2\sqrt{x}} - \frac {\sqrt{x}}{x} = \frac{\sqrt{x}}{2x}
$

The solution is supposed to be $\frac{\sqrt{x}(x-1)}{2x^2}$

2)
$
(ln \sqrt{\frac{a+x}{a-x}})' = \frac{a-x}{2(a+x)}
$

The solution is supposed to be $\frac{a}{a^2-x^2}$

2. #1) Think of it as $x+x^{-1}$ Use the power rule like normal.

3. if i use that i get $\frac{1}{2}x^{-\frac{1}{2}} - \frac{1}{2}x^{-\frac{3}{2}}$

4. Originally Posted by metlx
if i use that i get $\frac{1}{2}x^{-\frac{1}{2}} - \frac{1}{2}x^{-\frac{3}{2}}$
You shouldn't. d/dx x = 1 and d/dx $x^{-1}=-1(x)^{-2}$

5. Originally Posted by metlx
if i use that i get $\frac{1}{2}x^{-\frac{1}{2}} - \frac{1}{2}x^{-\frac{3}{2}}$
HOW do you get that? Please show your work. The power rule: $(x^n)'= nx^{n-1}$ won't give fractional powers if you don't start with them!

6. aww. i just realized I forgot the square root.

1) is $(\sqrt{x} + \frac{1}{\sqrt{x}})'$

7. Originally Posted by metlx
aww. i just realized I forgot the square root.

1) is $(\sqrt{x} + \frac{1}{\sqrt{x}})'$
Please read the replies you're getting. There should be NO SQUARE ROOTS unless you simplify after taking the derivative.

8. no i forgot to add the square root into the initial derivative..

I wrote $(x + \frac {1}{x})'$ instead of $(\sqrt{x} + \frac{1}{\sqrt{x}})'$

9. by the way, 2) is no longer needed. Managed to solve it

10. Then your first part if right. $\frac{d}{dx} \frac{1}{\sqrt{x}}= \frac{d}{dx} x^{-\frac{1}{2}}=-\frac{1}{2}x^{-\frac{3}{2}}$

11. I'm guessing that the solution $\frac{\sqrt{x}(x-1)}{2x^2}$ that is in the book is wrong then?

12. Originally Posted by metlx
I'm guessing that the solution $\frac{\sqrt{x}(x-1)}{2x^2}$ that is in the book is wrong then?
$\frac{1}{2\sqrt{x}}-\frac{1}{2}x^{-\frac{3}{2}}$

I am busy right now so I can't do this for you, but try to get these two terms into one expression. Then see if the book's answer makes more sense.