# Derivative of $\frac1{1-x}$

Why is this not correct: $$\frac{1}{1-x}= (1-x)^{-1}$$

now use chain-rule which gives: $(1-x)^{-2}$ times derivative of $(1-x)$ which is $-1$ so $$-1\cdot (1-x)^{-2}= \frac{-1}{(1-x)^2}$$

why is this incorrect?

Because if I use quotient rule on $1/(1-x)$ I get $$\frac{0 \cdot (1-x) - 1\cdot -1}{(1-x)^2}= \frac{1}{(1-x)^2}.$$

So why do I get with using chain rule on $(1-x)^{-1}$ a different answer? $$\frac{d}{dx} (1-x)^{-1}= \frac{d}{du} (u)^{-1} \cdot \frac{d}{dx} (u),$$ with $$u=1-x \Longrightarrow (u)^{-2}\cdot( -1)= \frac{-1}{(1-x)^2}$$

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It should be "which gives $(-1)\cdot(1-x)^{-2}\ldots$" – Git Gud Jul 7 '14 at 20:36

Remember $\frac{d}{dx}x^n=nx^{n-1}$, so in the first case you should have $$\frac{d}{dx}(1-x)^{-1}=(-1)(1-x)^{-2}\left(\frac{d}{dx}(1-x)\right)=(-1)(1-x)^{-2}(-1)=\frac{1}{(1-x)^2}$$
$$\frac d{dx} (1-x)^{-1} = -1\cdot (1 - x)^{-2} \cdot \underbrace{\frac{d}{dx}(1-x)}_{\large =\,-1}$$
$$= -1\cdot -1\cdot (1 - x)^{-2}= \frac{1}{(1-x)^2}$$