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Thread: [SOLVED] How to differentiate this?

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

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

    This is my take on the question:

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

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

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


    But the answer given was

    $\displaystyle \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 $\displaystyle \sqrt{1+2x}$, which is effectively multiplying by $\displaystyle 1$.

    To see it in action
    From:




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

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



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

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

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

    then, simplifying them, I get

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

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

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

    then, simplifying them, I get

    $\displaystyle =\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 :

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

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

    $\displaystyle
    \frac{(1+2x)(-2x)}{\sqrt{1+2x}} - \frac{1-x^2}{\sqrt{1 + 2x}}$
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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 $\displaystyle 1 + 2x$ instead of multiplying it with the denominator which is $\displaystyle \sqrt{1 + 2x}$. Oh...I feel kinda silly now. Not enough practice, I must say. Anyway, thanks for your time!
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