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Bear with me folks, I'm only learning.

How do I differentiate $y=\frac{1}{x}$ from first principles? I get to:

$\frac{dy}{dx}=\frac{y_2-y_1}{x_2-x_1}=\frac{\left(\frac{1}{x+h}\right)-\left(\frac{1}{x}\right)}{h}$

I then try to find a common denominator to subtract the terms in the numerator, but am then left with $\frac{dy}{dx}=\frac{-\frac{h}{x^2+xh}}{h}$ which always seems to reduce to zero (with $h$ tending to zero).

What's particularly frustrating is that I can easily find the answer ($-\frac{1}{x^2}$) using differentiation from rules. Any help enormously welcome.

FreeThought
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  • It doesn’t tend to 0. Cancel the $h$ first. – peek-a-boo Apr 25 '23 at 15:02
  • @peek-a-boo Thank you for your help, but alas your answer is completely beyond me. Why doesn't h tend to zero? When should I cancel it? – FreeThought Apr 25 '23 at 15:41
  • You have two fraction bars. Do some simple algebra and rewrite it with only one fraction bar. – peek-a-boo Apr 25 '23 at 16:18
  • @peek-a-boo Okay I need to divide everything in my final expression by h, which is equivalent to multiplying everything by 1/h. But this makes every term in my expression a factor of h. If h tends to zero (which as far as I understand, it must, if the difference between the two points is to be infinitely small), then I am left with 0/0, that is, zero! I know I am making some ridiculously stupid mistake here, but I just can't see it. – FreeThought Apr 25 '23 at 17:25
  • @peek-a-boo Figured it out - cancel out the hs before multiplying -h/h(x^2+xh) by 1/h. Avoids making everything a factor of h. I get it. Thank you. – FreeThought Apr 25 '23 at 19:40

1 Answers1

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The problem is unsolvable if one reasons as follows, even though the reasoning is sound:

$\frac{dy}{dx}=\frac{-\frac{h}{x^2+xh}}{h}$

$\longrightarrow\frac{dy}{dx}=-\frac{h}{x^2+xh}\div\frac{h}{1}$

$\longrightarrow\frac{dy}{dx}=-\frac{h}{x^2+xh}\times\frac{1}{h}$

$\longrightarrow\frac{dy}{dx}=-\frac{h}{h\left(x^2+xh\right)}$

The problem is that now everything is a factor of $h$ and so reduces to zero if $h$ tends to zero. Instead, one must reason as follows:

$\longrightarrow\frac{dy}{dx}=-\frac{h}{x^2+xh}\times\frac{1}{h}$

Cancel out the $h$s to give:

$\longrightarrow\frac{dy}{dx}=-\frac{1}{x^2+xh}\times\frac{1}{1}$

$\longrightarrow\frac{dy}{dx}=-\frac{1}{x^2+xh}$

If $h$ tends to zero this gives:

$\frac{dy}{dx}=-\frac{1}{x^2}$

FreeThought
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  • The first line of your post is misleading. The reasoning is sound and it does lead to the solution. You just have to keep going and have the presence of mind to cancel the $h$s – FShrike Apr 25 '23 at 20:19
  • @FShrike Agreed; reasoning which leads indirectly to an answer by way of a wrong answer can be equally as sound as reasoning which leads directly to an answer. – FreeThought Apr 25 '23 at 20:40