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Haven't touched Calculus in a very long time. Going back through it at the moment and this equation has me stumped right now:

$$ f(x)= \frac{1}{x}, \frac{f(x)-f(a)}{x-a} $$ I know how a regular differential quotient works but this is confusing to me as to what values go where. And my textbook only tells me the answer but doesn't show the steps to get to the answer. Any help would be much appreciated. I Googled for a bit looking for calculators that show every step and also looking for "unique differential quotients" and the like. But came out empty-handed. I'm not looking for an answer, I already have it. I just want to understand the process, what values get substituted and what not. Thanks.

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    $\begingroup$ You mean how to show it's $\dfrac{-1}{ax}$? $\endgroup$ Nov 15, 2020 at 23:15
  • $\begingroup$ @J.W.Tanner Yep, that's the answer in my textbook. $\endgroup$ Nov 15, 2020 at 23:17
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    $\begingroup$ I posted an answer. Let me know if you need more details $\endgroup$ Nov 15, 2020 at 23:18

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$$f'(a)=\lim\limits_{x\to a}\dfrac{\frac1x-\frac1a}{x-a}=\lim\limits_{x\to a}\dfrac{a-x}{ax}\dfrac{1}{x-a}=\lim\limits_{x\to a}\dfrac1{ax}\dfrac{a-x}{x-a}=\lim\limits_{x\to a}\dfrac{-1}{ax}=\dfrac{-1}{a^2}$$

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  • $\begingroup$ I don't understand how you got from a-x/ax * 1/x-a to -1/ax? $\endgroup$ Nov 15, 2020 at 23:26
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    $\begingroup$ I edited in an additional step. Is it clear now, or do you need further elucidation? $\endgroup$ Nov 15, 2020 at 23:28
  • $\begingroup$ That's perfect, seems I forgot some simple fraction operations. Lol, time to go back to square 1. Thanks. $\endgroup$ Nov 15, 2020 at 23:31

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