# Quotient rule of derivatives

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?

• 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 $\frac{2}{(1-x)^{3}}$. Jan 21, 2017 at 12:35
• i was computing Taylor series and i see that i filled directly value of x = 0 , which was a mistake, :( thanks for the answer :-) Jan 21, 2017 at 12:37
• Oh. Then I'm sorry.) Jan 21, 2017 at 12:39
• 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 $\frac{d}{dx}\left(\frac{1}{g(x)}\right)=-\frac{g'(x)}{g(x)^2}$ Jan 21, 2017 at 12:45
– amd
Jan 21, 2017 at 22:01

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

• It's a prime and not a second derivative of $1/(1-x)^2$ Jan 21, 2017 at 12:47
• @marwalix thanks. Jan 21, 2017 at 12:48

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

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

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.