I'm getting answer of $\frac{2}{(1-x)^3}$ , but online calculators suggest it's $\frac{-2}{(1-x)^3}$. I've tried it as $(1-x)^{-2}$, which results $-2*(1-x)*-1 = \frac{2}{(1-x)^3}$. Same result with $(\frac{f}{g})'=\frac{(f'*g-f*g')}{g^2}$ . Any help please?
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1Your answer is correct : $((1-x)^{-2})'=-2(1-x)^{-3}\cdot (1-x)'=2(1-x)^{-3}.$ – mathlove Apr 18 '15 at 13:02
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1Those calculators seem to be defective. – Michael Hoppe Apr 18 '15 at 13:04
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1Could you give the link to the calculator you used ? This could be very interesting for our community. Thanks. – Claude Leibovici Apr 18 '15 at 13:06
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WolframAlpha agrees with your answer, noting that $(1-x)^3 = -(x-1)^3$. Perhaps the calculator you used as well reversed the order of the $x$ and the $1$. – JMoravitz Apr 18 '15 at 13:08
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after the power and the chaine rule you will get $$-2(1-x)^{-3}(-1)$$
Dr. Sonnhard Graubner
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Let $f: x \mapsto 1 - x$ and $g: u \mapsto u^{-2}.$ Let $h: x \overset{f}{\mapsto} 1-x =: u \overset{g}{\mapsto} u^{-2}.$ Then $h'(x) = -2u^{-3}\big|_{u = 1-x}\cdot (-1) = 2(1-x)^{-3}.$
So yes, you are correct there.
Yes
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