# Derivative of $(1-x)^{-2}$

Wolfram Alpha is telling me that the answer is $$-\frac2{(x-1)^3}.$$ But I thought that by using the chain rule you multiply the front by $2$ then subtract the exponent by $1$ then multiply by the derivative of the inside ($-1$) to get $$\frac2{(1-x)^3}.$$ Wolfram Alpha is never wrong so please tell me what I'm missing its giving me a real headache.

• "Wolfram Alpha is never wrong" - it's a computer program; computer programs can certainly be wrong! Have you never seen a bug in software before? :) – Math1000 Mar 29 '15 at 0:19

hint: $(1-x)^{-3} = -(x-1)^{-3}$. This is the cause of your headache since the sign switched.

• lmao omg can't believe I didn't catch that. Totally didn't realize they simplified it like that thanks a lot. – jake Mar 29 '15 at 0:15
• Wolfram does this all the time. It's frustrating sometimes when it replaces e.g. $1-x$ with $-(x-1)$. – Math1000 Mar 29 '15 at 0:18

Let $f(x) = (1-x)^{-2}$.

Then $f'(x)= \frac{df(x)}{dx}= (-2) \times (-1) \times (1-x)^{-3} = \frac{2}{(1-x)^{3}}.$

And we are done.

Hint:

Just observe $-(1-x)^{-3}=(-1)^{-3}(1-x)^{-3}=[(-1)(1-x)]^{-3}=(x-1)^{-3}$

It sounds like your problem here isn't the chain rule. Your exponent is $-2$. The derivative of $u^{-2}=-2u^{-3}$. So if you have $u=1-x,\frac{du}{dx}$ is indeed $-1$, so you have $$\frac d{du}(u^{-2})\frac{du}{dx}=-2u^{-3}(-1)=2(1-x)^{-3}$$