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I'm watching the first MIT OCW 18.01 lecture. In this lecture, the lecturer gives a treatment of finding the derivative for $f(x) = \frac{1}{x}$. He presents this interstitial step $$\frac{\Delta f}{\Delta x} = \frac{\frac{1}{x_0+\Delta x} - \frac{1}{x_0}}{\Delta x}$$ and simplifies it to $$\frac{\Delta f}{\Delta x} = \frac{1}{\Delta x}\left(\frac{x_0 - \left(x_0 + \Delta x\right)}{\left(x_0 + \Delta x\right)x_0}\right)$$.

He explains the simplification as factoring out the $\frac{1}{\Delta x}$ from the denominator and then multiplying the resulting denominators $x_0 + \Delta x$ and $x_0$ to obtain the denominator $\frac{x_0 - \left(x_0 + \Delta x\right)}{\left(x_0 + \Delta x\right)x_0}$ and then "figuring out what the numerator had to be". It's not clear to me how the numerator in this case would be anything other than $-1$; indeed, after some simplification, he ends up with $\frac{\Delta f}{\Delta x} = \frac{-1}{x^2}$.

What are the steps involved in simplifying the numerator in this expression, and what am I missing here? To try to be more clear, I understand that ultimately the numerator cancels to be $-1$. I understand the power rule and how to use it to short-circuit round the step by step procedure. What I am not clear on is how, in the intermediate step, the numerator is an expression involving $x_0$ and $\Delta x$. How did this come up? Where does it come from?

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  • $\begingroup$ The result should be $-\frac{1}{x^2}$ $\endgroup$ – Peter Apr 6 '18 at 9:35
  • $\begingroup$ The derivative of $x^n$ is $nx^{n-1}$. Thus the derivetive of $\dfrac 1 x=x^{-1}$ must be: $(-1)x^{-2}=- \dfrac {1}{x^2}$. $\endgroup$ – Mauro ALLEGRANZA Apr 6 '18 at 9:39
  • $\begingroup$ Thank you for the correction, Peter, you are quite right. I have amended the question to show that. Mauro, I understand the power rule, what I'm asking about, specifically, is the $x_0 - \left(x_0 + \Delta x\right)$ numerator in the intermediate step. $\endgroup$ – Chris Apr 6 '18 at 9:48
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I am not completely clear on what you are asking, but the numerator $x_0-(x_0+\Delta x)$ comes from this: $$\frac{1}{x_0+\Delta x} - \frac{1}{x_0}=\frac{x_0}{x_0(x_0+\Delta x)}-\frac{x_0+\Delta x}{x_0(x_0+\Delta x)}=\frac{x_0-(x_0+\Delta x)}{x_0(x_0+\Delta x)}$$

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  • $\begingroup$ What I am asking is: how do you obtain the result in the middle? I understand the left side of the left $=$ and the right side of the right $=$ but not the middle. $\endgroup$ – Chris Apr 6 '18 at 10:12
  • $\begingroup$ @Chris For the fraction $\dfrac{1}{x_0+\Delta x}$, multiply both the numerator and denominator by $x_0$. We get $\dfrac{x_0}{x_0(x_0+\Delta x)}$, which is an equivalent fraction. Similar for the other fraction. $\endgroup$ – A. Goodier Apr 6 '18 at 10:15
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    $\begingroup$ OH. Oh. Right. Of course. Thank you! I was forgetting that bit. Now it makes sense. Thank you! $\endgroup$ – Chris Apr 6 '18 at 10:18
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we have $$\frac{1}{\Delta x}\left(\frac{x_0-x_0-\Delta x}{(x_0^2+\Delta xx_0}\right)=\frac{-1}{x_0^2+\Delta x x_0}$$ and for $\Delta x $ tends to $0%$ we get $$\frac{-1}{x_0^2}$$

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  • $\begingroup$ I understand how to get there, what I'm not sure is why the numerator in the intermediate step can be anything other than $-1$. $\endgroup$ – Chris Apr 6 '18 at 9:42
  • $\begingroup$ You can cancel the $\Delta x$ and $\frac{-\Delta x}{\Delta x}=-1$ $\endgroup$ – Dr. Sonnhard Graubner Apr 6 '18 at 9:44
  • $\begingroup$ Maybe I'm not being clear. I apologize! It's late here. I understand that you can cancel that way. What I don't understand is the intermediate step, where the numerator is $x_0 - \left(x_0 + \Delta x\right)$. This is the bit that's confusing me. Why is that not $-1$? $\endgroup$ – Chris Apr 6 '18 at 9:48
  • $\begingroup$ it is $$x_0-x_0-\Delta x=-\Delta x$$! $\endgroup$ – Dr. Sonnhard Graubner Apr 6 '18 at 9:49

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