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  1. #1
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    simple derivative of...

    could someone help me with this: if y= 1/(x+1) find y'...

    i tried and i got this equation x^2 +2x +1 is it correct???
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by juanfe_zodiac View Post
    could someone help me with this: if y= 1/(x+1) find y'...

    i tried and i got this equation x^2 +2x +1 is it correct???
    no

    write as y = (x + 1)^{-1}

    now use the chain rule
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  3. #3
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    simple derivative of...

    i still dont getting it...could u explain me...what chain is that? how do i know that the new derivative is correct?
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  4. #4
    Member RedBarchetta's Avatar
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    <br />
\begin{gathered}<br />
  y = \frac{1}<br />
{{x + 1}} \hfill \\<br />
  y' = ? \hfill \\ <br />
\end{gathered} <br />

    Use the quotient rule:

    <br />
\left( {\frac{u}<br />
{v}} \right)' = \frac{{u'v - uv'}}<br />
{{v^2 }}<br />

    Where u=1 and v=x+1.....
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by juanfe_zodiac View Post
    i still dont getting it...could u explain me...what chain is that? how do i know that the new derivative is correct?
    the chain rule says: \frac d{dx} f(g(x)) = f'(g(x)) \cdot g'(x)

    basically, to find the derivative of a composite function, differentiate the outer function (leaving the inner function intact) then multiply by the derivative of the inside function


    y = (x + 1)^{-1}

    \Rightarrow y' = -(x + 1)^{-2}(1) = \frac {-1}{(x + 1)^2}

    as for knowing the derivative is correct, look up the answer in your text ...or check it with integration, but i don't suppose you have started integration as yet
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