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I am working on this differetiation problem:

$ \frac{d}{dx}x(1-\frac{2}{x})$

and I am currently stuck at this point:

$1\cdot \left(1-\frac{2}{x}\right)+\frac{2}{x^2}x$

Symbolab tells me this simplifies to $1$ but I do not understand how. I am under the impression that;

$1\cdot \left(1-\frac{2}{x}\right)+\frac{2}{x^2}x \equiv 1- 2x^{-x^2}-2^{-x}$

  • 0
    what kind of reasoning did you use for your last line? that's a good one...^^ – 2017-02-03
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    i see: $n/x=(-)^n nx^{-x^n}$...you made my day (+1) – 2017-02-03
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    $2x \div (x \dot x) = 2x^{-x^2}$ – 2017-02-03
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    You are confusing $a/x=ax^{-1}$ and $a^{-x}$. – 2017-02-03

3 Answers 3

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$1\cdot \left(1-\frac{2}{x}\right)+\frac{2}{x^2}x $

= $1\cdot 1- 1 \cdot \frac{2}{x} + \frac{2}{x}$

= $1- \frac{2}{x} + \frac{2}{x}$

=$1$

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Do not get confused by fractions and exponents. You should remember that $\frac {k}{n} = kn^{-1}$ and not $k^{-n} $ and that $\frac {k}{n} l = kln^{-1}$ and not $kl^{-n} $.

We have $$1\times (1-\frac {2}{x}) = 1-\frac {2}{x} \tag {1}$$ and then $$\frac {2}{x^2}x = \frac {2}{x^2} \times x = \frac {2}{x} \tag {2} $$

What do we get by adding $(1)$ and $(2)$? The result is $1-\frac {2}{x} + \frac {2}{x} = 1$. Hope it helps.

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$$1\cdot (1-\frac2x) + \frac{2}{x^2}x = 1-\frac2x + \frac2x = 1$$ for all $x\neq 0$.

I have no idea where you got $x^{-x^2}$ from...

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    The $x^2$ was obviously taken from the denominator and scrambled with a $-1$ exponent. – 2017-02-03