First principle of differentiation

**99.95** · August 2nd 2010, 02:38 AM

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".

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**Chris L T521** · August 2nd 2010, 02:42 AM

Originally Posted by 99.95

[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".

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?

**99.95** · August 2nd 2010, 10:01 PM

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.

**eumyang** · August 3rd 2010, 05:44 AM

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

**99.95** · August 4th 2010, 02:30 AM

Originally Posted by eumyang

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

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?

**HallsofIvy** · August 4th 2010, 02:33 AM

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.

**99.95** · August 4th 2010, 02:37 AM

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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