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Power Rule and Quotient rule

  1. Mar 28, 2010 #1
    Hi everyone,
    I have been trying to do this problem in both ways but I cant get the same answer the book says. This is the problem:

    x/ sqrt (x^2 +1)

    With quotient rule I got until the point I have [(x^2 +1)^1/2 - x^2/(x^2 +1)^1/2]/(x^2 +1)
    And with power rule I have [1/sqrt(x^2 +1)] - [x^2/(x^2 +1)^3/2]

    If you guys can walk me through the problem, it would be nice. Thanks :wink:
     
  2. jcsd
  3. Mar 28, 2010 #2

    arildno

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    Well, let's look at your result from the quotient rule:
    [tex]\frac{\sqrt{x^{2}+1}-\frac{x^{2}}{\sqrt{x^{2}+1}}}{x^{2}+1}=\frac{\sqrt{x^{2}+1}}{x^{2}+1}-\frac{x^{2}}{(x^{2}+1)\sqrt{x^{2}+1}}=\frac{1}{\sqrt{x^{2}+1}}-\frac{x^{2}}{(x^{2}+1)^{\frac{3}{2}}}[/tex]

    Does that look familiar?

    Furthermore, we may expand our first fraction with the factor (x^{2}+1).
    Then, we get:

    [tex]\frac{(x^{2}+1)-x^{2}}{(x^{2}+1)^{\frac{3}{2}}}=\frac{1}{(x^{2}+1)^{\frac{3}{2}}}[/tex]
     
    Last edited: Mar 28, 2010
  4. Mar 28, 2010 #3
    Can you explain me in details how you combined that part of the fraction because I get lost here [tex]\frac{\sqrt {x^{2}+1}}{x^{2}+1}-\frac{x^{2}}{(x^{2}+1)\sqrt{x^{2}+1}}=\frac{1}{\sq rt{x^{2}+1}}-\frac{x^{2}}{(x^{2}+1)^{\frac{3}{2}}}[/tex].
     
    Last edited: Mar 28, 2010
  5. Mar 28, 2010 #4
    [tex]\frac{\sqrt {x^{2}+1}}{x^{2}+1}-\frac{x^{2}}{(x^{2}+1)\sqrt{x^{2}+1}}=\frac{1}{\sq rt{x^{2}+1}}-\frac{x^{2}}{(x^{2}+1)^{\frac{3}{2}}}[/tex]

    You first multiply so all terms have a common divisor

    [tex]\frac{\sqrt {x^{2}+1}*\sqrt{x^{2}+1}}{x^{2}+1}-\frac{x^{2}}{(x^{2}+1)\sqrt{x^{2}+1}}=\frac{1}{\sq rt{x^{2}+1}}-\frac{x^{2}}{(x^{2}+1)^{\frac{3}{2}}}[/tex]

    and then simplify

    as [tex]x^{2}+1[/tex] is the same as [tex](x^{2}+1)^{\frac{2}{2}[/tex] you multiply that with[tex](x^{2}+1)^{\frac{1}{2}[/tex]

    Hope this helps
     
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