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up vote 4 down vote favorite

I know this is really basic, but how do I differentiate this equation from first principles to find $\frac{dy}{dx}$:

$$ y = \frac{1}{x} $$

I tried this:

$$\begin{align} f'(x) = \frac{dy}{dx} & = \lim_{\delta x\to 0} \left[ \frac{f(x + \delta x) - f(x)}{\delta x} \right] \\ & = \lim_{\delta x\to 0} \left[ \frac{(x+\delta x)^{-1} - x^{-1}}{\delta x} \right] \\ & = \lim_{\delta x\to 0} \left[ \frac{1}{\delta x(x + \delta x )} - \frac{1}{x(\delta x)} \right] \\ & = \lim_{\delta x\to 0} \left[ \frac{1}{x(\delta x)} + \frac{1}{(\delta x)^2} - \frac{1}{x(\delta x)} \right] \\ & = 1 \end{align}$$

which is obviously wrong, since $f'(x) = - \frac{1}{x^2}$. Where am I going wrong?

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1  
how do you get from the third line to the fourth in your equation? –  Thomas yesterday
1  
Take the third line and reduce to same denominator. –  Claude Leibovici yesterday
3  
It doesn't necessarily hold that $\frac 1{a+b}=\frac 1a+\frac 1b$. –  Git Gud yesterday
    
Yes as @HartoSaarinen mentions, they are not equal. Thats where you went wrong. –  Tharindu yesterday

5 Answers 5

up vote 12 down vote accepted

$$\begin{align} f'(x) = \frac{dy}{dx} & = \lim_{\delta x\to 0} \left[ \frac{f(x + \delta x) - f(x)}{\delta x} \right] \\ & = \lim_{\delta x\to 0} \left[ \frac{(x+\delta x)^{-1} - x^{-1}}{\delta x} \right] \\ & = \lim_{\delta x\to 0} \left[\color{blue}{ \frac{1}{\delta x(x + \delta x )}} - \frac{1}{x(\delta x)} \right] \\ & = \lim_{\delta x\to 0} \left[ \color{blue}{\frac{1}{x(\delta x)} + \frac{1}{(\delta x)^2}} - \frac{1}{x(\delta x)} \right] \\ & = 1 \end{align}$$

Highlighted in blue is the mistaken step you took. Namely, you fell prey to the fallacy which is rarely true: $$\frac 1{a+b} = \frac 1a + \frac 1b\tag{fallacy}$$

The fallacy stems, I think, from the fact that we can split a sum when it's in the numerator, like thus, $$\frac {a+b}{c} = \frac ac + \frac bc$$ but we cannot do so when the sum is in the denominator.

What you can do is find the common denominator of the two terms (fractions) in the third line above, and then all should progress smoothly.

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up vote 3 down vote

$$\lim_{\delta x\to 0} \left[ \frac{1}{\delta x(x + \delta x )} - \frac{1}{x(\delta x)} \right]=\lim_{\delta x\to 0} \left[\frac{-\delta x-x}{x\delta x (\delta x+x)}+\frac{x}{x\delta x (\delta x+x)} \right ]=\lim_{\delta x\to 0} \left[-\frac{\delta x}{x\delta x(x+\delta x)} \right]=\lim_{\delta x\to 0} \left[\frac{-1}{x(x+\delta x)}\right ]=\lim_{\delta x\to 0} \left[\frac{-1}{x\delta x+x^2}\right ]=-\frac{1}{x^2}$$

This is as detailed as it can go I think.

Remember that $$\frac{1}{a+b} \neq \frac1a +\frac1b, \forall a,b$$

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up vote 2 down vote

You did some strange calculation to arrive at the fourth line. I prefer to use $h$ instead of $\delta x$. Then $$\frac{1}{h}\left( \frac{1}{x+h} - \frac{1}{x}\right) = \frac{1}{h} \frac{x-x-h}{x(x+h)} = \frac{-1}{x(x+h)} $$ which does what you are expecting.

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up vote 1 down vote

$$\begin{align} f(x+h)-f(x) &= \dfrac{1}{x+h} - \dfrac{1}{x} \\ &= \dfrac{x-h-x}{x(x+h)} \\ &= \dfrac{-h}{x(x+h)} \end{align}$$

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up vote 1 down vote

$$f^{´}(x) = \lim_{h\to 0} \displaystyle \frac{f(x+h)-f(x)}{h} =$$

$$\lim_{h\to 0}\displaystyle \frac{\frac{1}{x+h}-\frac{1}{x}}{h} = \lim_{h\to 0}\displaystyle \frac{\frac{x-x-h}{x(x+h)}}{h} = \lim_{h\to 0}\displaystyle \frac{\frac{-h}{x(x+h)}}{h} = \lim_{h\to 0}\displaystyle \frac{-h}{hx(x+h)}$$

$$ \lim_{h\to 0}\displaystyle \frac{-1}{x^{2}+xh}= \frac{-1}{x^{2}} $$

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