1. ## Derivitives

Find f '(a).

$\displaystyle f(x) = \frac{1}{\sqrt{x+6}}$

plugging this into the formula i get

$\displaystyle \lim_{h \to 0} \frac{\frac{1}{\sqrt{x+h+6}}-\frac{1}{\sqrt{x+6}}}{h}$

which simplified all the way down to

$\displaystyle \lim_{h \to 0} \frac{-h}{h(\sqrt{x+h+6})(\sqrt{x+6})(\sqrt{x+6}+\sqrt{x+ h+6})}$

then to

$\displaystyle \lim_{h \to 0} \frac{-1}{(\sqrt{x+h+6})(\sqrt{x+6})(\sqrt{x+6}+\sqrt{x+h +g})}$

and finally

$\displaystyle \lim_{h \to 0} \frac{-1}{(x+6)^2}$

but...its wrong

any help?

EDIT: or would the final answer be

$\displaystyle \lim_{h \to 0} \frac{-1}{(x+6)(\sqrt{x+6}+\sqrt{x+6}})$

2. Hello,
Originally Posted by silencecloak
Find f '(a).

$\displaystyle f(x) = \frac{1}{\sqrt{x+6}}$

plugging this into the formula i get

$\displaystyle \lim_{h \to 0} \frac{\frac{1}{\sqrt{x+h+6}}-\frac{1}{\sqrt{x+6}}}{h}$

which simplified all the way down to

$\displaystyle \lim_{h \to 0} \frac{-h}{h(\sqrt{x+h+6})(\sqrt{x+6})(\sqrt{x+6}+\sqrt{x+ h+6})}$

then to

$\displaystyle \lim_{h \to 0} \frac{-1}{(\sqrt{x+h+6})(\sqrt{x+6})(\sqrt{x+6}+\sqrt{x+h +6})}$
Correct !

and finally

$\displaystyle \lim_{h \to 0} \frac{-1}{(x+6)^2}$

but...its wrong

any help?
I see where the problem is

When substituting h=0, we get :
$\displaystyle \frac{-1}{(\sqrt{x+6})(\sqrt{x+6})(\sqrt{x+6} {\color{red}+} \sqrt{x+6})}$

$\displaystyle (\sqrt{x+6})(\sqrt{x+6})=x+6$, you got it right !

But $\displaystyle \sqrt{x+6} {\color{red}+} \sqrt{x+6}={\color{red}2} \sqrt{x+6}$

Got it ?

3. Originally Posted by Moo
Hello,

Correct !

I see where the problem is

When substituting h=0, we get :
$\displaystyle \frac{-1}{(\sqrt{x+6})(\sqrt{x+6})(\sqrt{x+6} {\color{red}+} \sqrt{x+6})}$

$\displaystyle (\sqrt{x+6})(\sqrt{x+6})=x+6$, you got it right !

But $\displaystyle \sqrt{x+6} {\color{red}+} \sqrt{x+6}={\color{red}2} \sqrt{x+6}$

Got it ?
darn algebra, so easy to get mixed up

so i end up with

$\displaystyle \lim_{h \to 0} \frac{-1}{(x+6)(2\sqrt{x+6})}$

ah, but since its for $\displaystyle f'(a)$, do i now need to replace the x with a?

4. Originally Posted by silencecloak
darn algebra, so easy to get mixed up

so i end up with

$\displaystyle \lim_{h \to 0} \frac{-1}{(x+6)(2\sqrt{x+6})}$

ah, but since its for $\displaystyle f'(a)$, do i now need to replace the x with a?
Yes =)
And it is correct now !

5. awesome

thank you very much, as usual!