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Math Help - [SOLVED] How to differentiate this?

  1. #1
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    [SOLVED] How to differentiate this?

    \frac{1-x^2}{\sqrt{1 + 2x}}

    This is my take on the question:

    =\frac{\sqrt{1+2x}(-2x) - (1-x^2)\frac{1}{\sqrt{1 + 2x}}}{1 + 2x}

    =\frac{(-2x)(1 + 2x) - (1 - x^2)}{\sqrt{1 + 2x}} \times \frac{1}{1 + 2x}

     = \frac{-2x - 1 + x^2}{\sqrt{1 + 2x}}


    But the answer given was

     \frac{- (3x^2 + 2x + 1)}{(\sqrt{1 + 2x})^3}

    Can you tell me where I have gone wrong? Thanks.
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  2. #2
    Senior Member I-Think's Avatar
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    Mistake

    Good night(or morning or afternoon, depending on your location)

    If I understand your work correctly, then this is your mistake



    By doing this, you change the expression completely.

    What you should have done is multiply the numerator and denominator by \sqrt{1+2x}, which is effectively multiplying by 1.

    To see it in action
    From:




    *(\frac{\sqrt{1+2x}}{\sqrt{1+2x}})

    =\frac{-2x-4x^2-1+x^2}{\sqrt{{1+2x}}^3}



    And there you have it
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  3. #3
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    Hmm...okay but what I actually did was this:


    from
    =\frac{\sqrt{1+2x}(-2x) - (1-x^2)\frac{1}{\sqrt{1 + 2x}}}{1 + 2x}

    I equated the denominator of the numerator of the above to \sqrt{1 + 2x} to become

    =\frac{\frac{(\sqrt{1 + 2x})(-2x)(\sqrt{1 + 2x}) - (1 - x^2)}{\sqrt{1 + 2x}}}{1 + 2x}

    then, simplifying them, I get

    =\frac{(-2x)(1 + 2x) - (1 - x^2)}{\sqrt{1 + 2x}} \times \frac{1}{1 + 2x}

    Is it against the math logic if I equate the denominator of the fractions in the numerator of the expression?
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  4. #4
    MHF Contributor
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    Quote Originally Posted by mark1950 View Post
    Hmm...okay but what I actually did was this:


    from
    =\frac{\sqrt{1+2x}(-2x) - (1-x^2)\frac{1}{\sqrt{1 + 2x}}}{1 + 2x}

    I equated the denominator of the numerator of the above to \sqrt{1 + 2x} to become

    =\frac{\frac{(\sqrt{1 + 2x})(-2x)(\sqrt{1 + 2x}) - (1 - x^2)}{\sqrt{1 + 2x}}}{1 + 2x}

    then, simplifying them, I get

    =\frac{(-2x)(1 + 2x) - (1 - x^2)}{\sqrt{1 + 2x}} \times \frac{1}{1 + 2x}

    Is it against the math logic if I equate the denominator of the fractions in the numerator of the expression?

    HI Mark , you are not wrong . Just that you worked it out wrongly .

    For the numerator :

     <br />
\sqrt{1+2x}(-2x) - (1-x^2)\frac{1}{\sqrt{1 + 2x}}

     <br />
\sqrt{1+2x}(-2x) - \frac{1-x^2}{\sqrt{1 + 2x}}

     <br />
\frac{(1+2x)(-2x)}{\sqrt{1+2x}} - \frac{1-x^2}{\sqrt{1 + 2x}}
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  5. #5
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    Huh? What's the difference? If I simplify your numerator, wouldn't it be the same as my numerator?
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  6. #6
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    Hey, wait...I think I know the problem now! Yes, I mistakenly canceled out the 1 + 2x instead of multiplying it with the denominator which is \sqrt{1 + 2x}. Oh...I feel kinda silly now. Not enough practice, I must say. Anyway, thanks for your time!
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