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Math Help - Derivatives involving both product/quotient rule

  1. #1
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    Derivatives involving both product/quotient rule

    Find f'(x) for y = (x+5 / x+1) (2x+1)

    I don't know how to proceed here. Do I use the quotient rule inside the first brackets? Then what?
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  2. #2
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    First thing you can do is use the product rule
     y' = \frac{d}{dx}\left[\frac{x+5}{x+1}\right] \cdot (2x+1) + 2\cdot<br />
\frac{x+5}{x+1}.
    From here we can take the derivative using the quotient rule on \frac{d}{dx}\left[\frac{x+5}{x+1}\right] = \frac{(x+1) - (x + 5)}{(x+1)^2}.
    Combine this with the previous result
     y' = \frac{(x+1) - (x + 5)}{(x+1)^2} \cdot (2x+1) + 2\cdot<br />
\frac{x+5}{x+1}.
    Now simplify.
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  3. #3
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    Quote Originally Posted by lvleph View Post
    First thing you can do is use the product rule
     y' = \frac{d}{dx}\left[\frac{x+5}{x+1}\right] \cdot (2x+1) + 2\cdot<br />
\frac{x+5}{x+1}.
    From here we can take the derivative using the quotient rule on \frac{d}{dx}\left[\frac{x+5}{x+1}\right] = \frac{(x+1) - (x + 5)}{(x+1)^2}.
    Combine this with the previous result
     y' = \frac{(x+1) - (x + 5)}{(x+1)^2} \cdot (2x+1) + 2\cdot<br />
\frac{x+5}{x+1}.
    Now simplify.
    You don't end up with 2(x^2 + 2x + 3) / (x+1)^2 do you?
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  4. #4
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    Yes, that is what www.wolframalpha.com gives.
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