# Math Help - Differentiation - First Principles

1. ## Differentiation - First Principles

Find the derivative of 1 / square root of 2x

from first principles.

hellp !

xx

2. 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, $f(x) = \frac 1{\sqrt{2x}}$

3. 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

4. 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?

5. 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 ?

6. 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 ?
$\frac {\frac 1{\sqrt{2(x + h)}} - \frac 1{\sqrt{2x}}}h = \frac {\frac {\sqrt{2x} - \sqrt{2(x + h)}}{\sqrt{2(x + h)} \sqrt{2x}}}h = \frac {\sqrt{2x} - \sqrt{2(x + h)}}{h( \sqrt{2(x + h)} \sqrt{2x})} = \frac {\sqrt{2x} - \sqrt{2(x + h)}}{h(\sqrt{4x(x + h)})}$

7. 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 ?

8. 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

9. how do you get h out of the denominator then ?

10. Originally Posted by Jen1603
how do you get h out of the denominator then ?
did you simplify the numerator?

11. 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

12. 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 = ...

13. 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 ?

14. 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...?

15. 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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