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It gives me $\frac{1}{1-x}$ instead of $\frac{-1}{1-x}$. I tried arranging the parens in every possible way I could think of. I used the command Delvar $x$ to make sure there's no modifications to that variable. I tried radian and degree modes.

At this point, is my only option to reset it?

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    Maybe we can figure this out by trying some other functions. What comes out if you differentiate $\ln(x-1)$? What about $\ln(\cos(x))$? – 2017-01-26
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    1/x-1 and -tan(x) – 2017-01-26
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    That all seems fine... what about $\ln(-x)$? – 2017-01-26
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    fine why? 1/x-1 is not correct... – 2017-01-26
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    The derivative of $\ln(x-1)$ is indeed $1/(x-1)$ – 2017-01-26
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    it's not. It's -1/(x-1) – 2017-01-26
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    Applying the chain rule, we have $$ \frac{d}{dx}\ln(x-1) = \frac 1{x-1} \cdot \frac{d}{dx}[x-1] = \frac 1{x-1} $$ – 2017-01-26

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