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Quotient rule of derivative is: $(\frac{f}{g})^{\prime}$ = $(\frac{f^{\prime}g - g^{\prime}f}{g^2})$ but when I compute a deriative of $\frac{1}{(1-x)}$ , it gives $\frac{1}{(1-x)^2}$ which is right but taking second derivative of this gives me $\frac{2}{(1-x)^4}$ but the right answer is $\frac{2}{(1-x)^3}$. Where do I make a mistake?

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    How do you get $\frac{2}{(1-x)^{4}}$? If we take $f=1$ and $g=(1-x)^{2}$. Than as in previous $f'=0$, $g'=-2(1-x)$. So you'll have $\frac{-(1\cdot (-2) \cdot(1-x))}{(1-x)^{4}}$. Which will yield $\frac{2}{(1-x)^{3}}$. – 2017-01-21
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    i was computing Taylor series and i see that i filled directly value of x = 0 , which was a mistake, :( thanks for the answer :-) – 2017-01-21
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    Oh. Then I'm sorry.) – 2017-01-21
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    I think that using the formula you mentioned to compute the successive derivatives of $x\mapsto\frac{1}{1-x}$ is not the best point of view. It's easier if you consider using $\frac{d}{dx}\left(\frac{1}{g(x)}\right)=-\frac{g'(x)}{g(x)^2}$ – 2017-01-21
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    When asking for someone to point out where you made a mistake, show your work—we’re not all mind-readers. – 2017-01-21

3 Answers 3

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Let's derive

$$\left({1\over (1-x)^2}\right)^{'}={(1)'\cdot (1-x)^2-1\cdot\left((1-x)^2\right)^{'}\over (1-x)^4}$$

Now $(1)'=0$, $\left((1-x)^2\right)^{'}=-2(1-x)$ and so the derivative rewrites as

$$\left({1\over (1-x)^2}\right)^{'}={2(1-x)\over (1-x)^4}={2\over (1-x)^3}$$

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You get this because you can cancel out one $(1-x)$ in the denominator.

$\Big(\frac{1}{(1-x)^2}\Big)'=\frac{0*((1-x)^2)'-1*((1-x)^2)'}{((1-x)^2)^2}=\frac{2(1-x)}{(1-x)^4}=\frac{2}{(1-x)^3}$.

If you just follow the formula you had carefully, with $f=1$ and $g=(1-x)^2$, this is what you get.

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$\require{cancel}$ $\Big(\dfrac{1}{(1-z)^2}\Big)'=\dfrac{0\times(1-z)^2-2(-1)(1-z)\times1}{(1-z)^4}=\dfrac{2(1-z)}{(1-z)^4}=\dfrac{2‎‏\cancel{(1-z)}}{(1-z)^\cancel{4}}=\dfrac{2}{(1-z)^3}$

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    It's a prime and not a second derivative of $1/(1-x)^2$ – 2017-01-21
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    @marwalix thanks. – 2017-01-21