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Math Help - First principle of differentiation

  1. #1
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    First principle of differentiation

    Hi, just started differentiation topic a few weeks ago.
    My question is the attached image, I would appreciate if you provided me with procedures.
    Thanks
    (Originally I thought this was a limits question, however, i am unsure because in subbing in 1/x i kept getting "0".
    First principle of differentiation-untitled.jpgClick image for larger version. 

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    Last edited by 99.95; August 4th 2010 at 02:35 AM.
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by 99.95 View Post
    [IMG]file:///C:/DOCUME%7E1/YASHMI%7E1/LOCALS%7E1/Temp/moz-screenshot.png[/IMG]Hi, just started differentiation topic a few weeks ago.
    My question is the attached image, I would appreciate if you provided me with procedures.
    Thanks
    (Originally I thought this was a limits question, however, i am unsure because in subbing in 1/x i kept getting "0".
    Click image for larger version. 

Name:	untitled.JPG 
Views:	71 
Size:	16.0 KB 
ID:	18403Click image for larger version. 

Name:	untitled.JPG 
Views:	71 
Size:	16.0 KB 
ID:	18403
    You want to find the difference quotient given that f(x)=\dfrac{1}{x}. It follows that f(x+h)=\dfrac{1}{x+h}.

    Therefore,
    \begin{aligned}\dfrac{f(x+h)-f(x)}{h}&=\dfrac{\frac{1}{x+h}-\frac{1}{x}}{h}\\&=\dfrac{x-(x+h)}{hx(x+h)}\\&=\ldots\end{aligned}.

    I leave it for you to finish the problem. Does this make sense though?
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  3. #3
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    Oh, i see, so you subbed in 1/x and then used quotient rule?
    your second step though, when you obtain x - (x + h)
    is that the result of getting the reciprocal of 1/x+h - 1/x ?
    Thanks.
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  4. #4
    Senior Member eumyang's Avatar
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    That was the result of multiplying by the LCD of the two denominators in the numerator of the complex fraction. The LCD would be x(x + h), so
    \begin{aligned}<br />
\dfrac{f(x+h)-f(x)}{h}&=\dfrac{\frac{1}{x+h}-\frac{1}{x}}{h}\\<br />
&= \dfrac{\frac{1}{x+h}-\frac{1}{x}}{h} \cdot \dfrac{x(x + h)}{x(x +h)} \\<br />
&= \dfrac{\frac{x(x + h)}{x+h}-\frac{x(x + h)}{x}}{hx(x + h)} \\<br />
&=\dfrac{x-(x+h)}{hx(x+h)}\\<br />
&=\ldots\end{aligned}
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  5. #5
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    Quote Originally Posted by eumyang View Post
    That was the result of multiplying by the LCD of the two denominators in the numerator of the complex fraction. The LCD would be x(x + h), so
    \begin{aligned}<br />
\dfrac{f(x+h)-f(x)}{h}&=\dfrac{\frac{1}{x+h}-\frac{1}{x}}{h}\\<br />
&= \dfrac{\frac{1}{x+h}-\frac{1}{x}}{h} \cdot \dfrac{x(x + h)}{x(x +h)} \\<br />
&= \dfrac{\frac{x(x + h)}{x+h}-\frac{x(x + h)}{x}}{hx(x + h)} \\<br />
&=\dfrac{x-(x+h)}{hx(x+h)}\\<br />
&=\ldots\end{aligned}
    heh, thanks, embarrassing to see that year 9 maths has let me down (simplifying complex fractions
    Just confirming, in the final case you present me,
    I use y = u(x)/v(x), then y'= u'(x)v(x) - u(x)v'(x) / [v(x)]^2 (The quotient rule)
    Correct?
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  6. #6
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    You certainly can use the quotient rule to differentiate a quotient like 1/x but you said, in your first post, that you were to use the "limit of the difference quotient" definition of the derviative.
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  7. #7
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    Oh i'm once again embarrassed, I started this topic 3 days ago, so i am not so great with terminology. Is this when h--> 0
    and i sub in h--> after further factorising?
    thanks
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