# Thread: First principle of differentiation

1. ## 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".

2. 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?

3. 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.

4. 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}

5. 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?

6. 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.

7. 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