# Differentiation - First Principles

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• Nov 18th 2008, 06:12 AM
Jen1603
Differentiation - First Principles
Find the derivative of 1 / square root of 2x

from first principles.

hellp !

xx
• Nov 18th 2008, 07:45 AM
Jhevon
Quote:

Originally Posted by Jen1603
Find the derivative of 1 / square root of 2x

from first principles.

hellp !

xx

when you see "from first principles" they are talking about using the definition of the derivative.

there are two (equivalent) definitions:

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

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

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

alternatively, you may use the definition

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

in this case you are required to find the limit

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

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

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

here of course, $f(x) = \frac 1{\sqrt{2x}}$
• Nov 19th 2008, 04:54 AM
Jen1603
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, 07:48 AM
Jhevon
Quote:

Originally Posted by Jen1603
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

combine the fractions in the numerator and simplify so that you have one fraction overall. then rationalize the numerator of the new fraction by multiplying by the conjugate over itself. can you finish now?
• Nov 19th 2008, 10:51 AM
Jen1603
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, 03:35 PM
Jhevon
Quote:

Originally Posted by Jen1603
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 ?

$\frac {\frac 1{\sqrt{2(x + h)}} - \frac 1{\sqrt{2x}}}h = \frac {\frac {\sqrt{2x} - \sqrt{2(x + h)}}{\sqrt{2(x + h)} \sqrt{2x}}}h = \frac {\sqrt{2x} - \sqrt{2(x + h)}}{h( \sqrt{2(x + h)} \sqrt{2x})} = \frac {\sqrt{2x} - \sqrt{2(x + h)}}{h(\sqrt{4x(x + h)})}$
• Nov 19th 2008, 03:43 PM
Jen1603
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, 03:45 PM
Jhevon
Quote:

Originally Posted by Jen1603
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 ?

don't worry about simplifying the denominator yet. just leave it in its factorized form.

simplify the numerator, and something should cancel which would allow you to take the limit, which you can't do yet since h is in the denominator
• Nov 19th 2008, 03:46 PM
Jen1603
how do you get h out of the denominator then ?
• Nov 19th 2008, 04:13 PM
Jhevon
Quote:

Originally Posted by Jen1603
how do you get h out of the denominator then ?

did you simplify the numerator?
• Nov 19th 2008, 04:16 PM
Jen1603
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, 04:18 PM
Jhevon
Quote:

Originally Posted by Jen1603
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

that is not simplifying the numerator

2x - 2(x + h) = 2x - 2x - 2h = ...
• Nov 19th 2008, 04:20 PM
Jen1603
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, 04:23 PM
Jhevon
Quote:

Originally Posted by Jen1603
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 ?

...can't you cancel the h's...?
• Nov 19th 2008, 04:25 PM
Jen1603
cancel the h with the h on the denominator ?

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