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Math Help - Differentiation - First Principles

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    Differentiation - First Principles

    Find the derivative of 1 / square root of 2x

    from first principles.

    hellp !

    xx
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    Quote Originally Posted by Jen1603 View Post
    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}}
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    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
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    Quote Originally Posted by Jen1603 View Post
    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?
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    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 ?
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    Quote Originally Posted by Jen1603 View Post
    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)})}
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    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 ?
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    Quote Originally Posted by Jen1603 View Post
    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
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    how do you get h out of the denominator then ?
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    Quote Originally Posted by Jen1603 View Post
    how do you get h out of the denominator then ?
    did you simplify the numerator?
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    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
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    Quote Originally Posted by Jen1603 View Post
    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 = ...
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    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 ?
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    Quote Originally Posted by Jen1603 View Post
    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...?
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    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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