Find the derivative of 1 / square root of 2x

from first principles.

hellp !

xx

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- Nov 18th 2008, 05:12 AMJen1603Differentiation - First Principles
Find the derivative of 1 / square root of 2x

from first principles.

hellp !

xx - Nov 18th 2008, 06:45 AMJhevon
when you see "from first principles" they are talking about using the definition of the derivative.

there are two (equivalent) definitions:

$\displaystyle f'(x) = \lim_{h \to 0} \frac {f(x + h) - f(x)}h$

if this is the definition you have been taught, then you are required to find the limit

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

alternatively, you may use the definition

$\displaystyle f'(a) = \lim_{x \to a} \frac {f(x) - f(a)}{x - a}$

in this case you are required to find the limit

$\displaystyle f'(a) = \lim_{x \to a} \frac { \frac 1{\sqrt{2x}} - \frac 1{\sqrt{2a}}}{x - a}$

(and replace $\displaystyle a$ with $\displaystyle x$ when finished)

Notation: $\displaystyle f'(x)$ and $\displaystyle f'(a)$ mean "the derivative of $\displaystyle f(x)$ (with respect to $\displaystyle x$)" and "the derivative of $\displaystyle f(x)$ (with respect to x) at the point $\displaystyle x = a$" respectively.

here of course, $\displaystyle f(x) = \frac 1{\sqrt{2x}}$ - Nov 19th 2008, 03:54 AMJen1603
I got as far as the first f '(x) equation but i wasnt sure how to simplify it with the two fractions and the square roots

- Nov 19th 2008, 06:48 AMJhevon
- Nov 19th 2008, 09:51 AMJen1603
i had tried multiply by the conjugate , but firstly im unsure how to combine the fractions and therefore it all gets a bit messy. could you show me ?

- Nov 19th 2008, 02:35 PMJhevon
$\displaystyle \frac {\frac 1{\sqrt{2(x + h)}} - \frac 1{\sqrt{2x}}}h = \frac {\frac {\sqrt{2x} - \sqrt{2(x + h)}}{\sqrt{2(x + h)} \sqrt{2x}}}h = \frac {\sqrt{2x} - \sqrt{2(x + h)}}{h( \sqrt{2(x + h)} \sqrt{2x})} = \frac {\sqrt{2x} - \sqrt{2(x + h)}}{h(\sqrt{4x(x + h)})}$

- Nov 19th 2008, 02:43 PMJen1603
Thanks , so then i multiplyed by the conjugate .

so ive ended up with 2x - 2(x + h) on the top , but a really mess denominator. how do i simplify that ? - Nov 19th 2008, 02:45 PMJhevon
- Nov 19th 2008, 02:46 PMJen1603
how do you get h out of the denominator then ?

- Nov 19th 2008, 03:13 PMJhevon
- Nov 19th 2008, 03:16 PMJen1603
yeah , well i think i did = /

i multiplyed it by its conjugate over itself

and got 2x - 2 (x + h)

on the top , and then a mess on the bottom. and now im not sure what to do - Nov 19th 2008, 03:18 PMJhevon
- Nov 19th 2008, 03:20 PMJen1603
well i was getting confused about that aswell

because the x's cancel , so your left with -2h ...but then its as h tends to 0 , so would that not make the numerator zero ? - Nov 19th 2008, 03:23 PMJhevon
- Nov 19th 2008, 03:25 PMJen1603
cancel the h with the h on the denominator ?

but what about the h's that are inside square roots on the denominator ?