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Math Help - Very basic derivatives: 1/(1-x) and x/(1-x)

  1. #1
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    Question Very basic derivatives: 1/(1-x) and x/(1-x)

    I don't know what I'm doing wrong (but I know I'm doing something) but I keep getting these two derivatives the same:
    <br />
(\tfrac{1}{1-x})'=\tfrac{-1}{(1-x)^2}(-1)=\tfrac{1}{(1-x)^2}<br />

    <br />
(\tfrac{x}{1-x})'=\tfrac{x'(1-x)-x(1-x)'}{(1-x)^2}=\tfrac{1-x+x}{(1-x)^2}=\tfrac{1}{(1-x)^2}<br />
    But two derivatives can be the same only if the functions differ only with a constant!

    I know, I must be making some terribly stupid mistake somewhere. But I can't find it.
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  2. #2
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    Quote Originally Posted by lampak View Post
    I don't know what I'm doing wrong (but I know I'm doing something) but I keep getting these two derivatives the same:
    <br />
(\tfrac{1}{1-x})'=\tfrac{-1}{(1-x)^2}(-1)=\tfrac{1}{(1-x)^2}<br />

    <br />
(\tfrac{x}{1-x})'=\tfrac{x'(1-x)-x(1-x)'}{(1-x)^2}=\tfrac{1-x+x}{(1-x)^2}=\tfrac{1}{(1-x)^2}<br />
    But two derivatives can be the same only if the functions differ only with a constant!

    I know, I must be making some terribly stupid mistake somewhere. But I can't find it.
    \frac{1}{1-x}-1 = \frac{1}{1-x}-\frac{1-x}{1-x}=\frac{x}{1-x}
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  3. #3
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    Quote Originally Posted by lampak View Post
    I don't know what I'm doing wrong (but I know I'm doing something) but I keep getting these two derivatives the same:
    <br />
(\tfrac{1}{1-x})'=\tfrac{-1}{(1-x)^2}(-1)=\tfrac{1}{(1-x)^2}<br />

    <br />
(\tfrac{x}{1-x})'=\tfrac{x'(1-x)-x(1-x)'}{(1-x)^2}=\tfrac{1-x+x}{(1-x)^2}=\tfrac{1}{(1-x)^2}<br />
    But two derivatives can be the same only if the functions differ only with a constant!

    I know, I must be making some terribly stupid mistake somewhere. But I can't find it.
    \frac{1}{1-x} and \frac{x}{1-x} differ by a constant.

    \frac{x}{1-x} = \frac{x+1-1}{1-x} = \frac{1 - (1-x)}{1-x} = \frac{1}{1-x} - 1
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    Great thanks for quick reply
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