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Thread: Basic calculus Quotient rule question

  1. #1
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    Basic calculus Quotient rule question

    Hi iam doin quotient rules in calculus and iam just getting stuck at a certain stage in this question.
    The question is
    $\displaystyle = \frac{3x+1}{\sqrt{x+1}}$
    and the answer they have down is
    $\displaystyle =\frac{3x+5}{2\sqrt{(x+1)^3}}$
    and iam using the forumula
    $\displaystyle y=\frac{vu'-uv'}{v^2}$
    and iam getting stuck at
    $\displaystyle \frac{3(x+1)^\frac{1}{2}-\frac{1}{2}(x+1)^{-\frac{1}{2}}(3x+1)}{x+1}$
    could someone please show me the working out and how you do this question thx
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  2. #2
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    Multiply tnumerator and denominator by $\displaystyle 2(x+1)^{1/2}$ and note that $\displaystyle \sqrt{(x+1)^3}=(x+1)^{3/2}. $
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  3. #3
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    Still a little confused would anybody be able to show me the working out steps please. Whipflip i can kind of see how that gets the right answer but why do you do that?
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  4. #4
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    $\displaystyle \frac{\displaystyle 3(x+1)^{\frac{1}{2}} \ - \ \frac{1}{2} \cdot \frac{1}{(x+1)^{\frac{1}{2}}} \cdot (3x+1)}{(x+1)^1}$

    Notice that if we follow whipflip's advice, multiplying through with $\displaystyle \frac{(x+1)^{\frac{1}{2}}}{(x+1)^{\frac{1}{2}}}$ will give us that $\displaystyle (x+1)^{\frac{3}{2}}$ we wanted in the denominator:
    $\displaystyle \frac{\displaystyle 3(x+1)^{\frac{1}{2}} \ - \ \frac{1}{2} \cdot \frac{1}{(x+1)^{\frac{1}{2}}} \cdot (3x+1)}{(x+1)^1} \cdot {\color{blue}\frac{(x+1)^{\frac{1}{2}}}{(x+1)^{\fr ac{1}{2}}}}$

    $\displaystyle = \frac{\displaystyle 3(x+1)^{\frac{1}{2}}{\color{blue}(x+1)^{\frac{1}{2 }}} - \frac{1}{2} \cdot \frac{{\color{blue}(x+1)^{\frac{1}{2}}}}{(x+1)^{\f rac{1}{2}}} \cdot (3x+1)}{(x+1)^1{\color{blue}(x+1)^{\frac{1}{2}}}}$ .... (Simply multiplied in the blue factor)

    $\displaystyle = \frac{\displaystyle 3(x+1)^1 - \frac{1}{2} \cdot {\color{blue}(1)} \cdot (3x+1)}{(x+1)^{\frac{3}{2}}}$ .... (Used the fact that: $\displaystyle a^na^m = a^{n+m}$ and simplified)

    $\displaystyle = \cdots$
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