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Math Help - Derivative algebra

  1. #1
    Junior Member Greymalkin's Avatar
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    Derivative algebra

    Problem below:
    Derivative algebra-trig2.png
    Explain to me how \frac{1}{2}\sqrt{-x}(-1)-1 becomes \frac{-1}{2\sqrt{-x}}-1
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  2. #2
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    Re: Derivative algebra

    Quote Originally Posted by Greymalkin View Post
    Problem below:
    Click image for larger version. 

Name:	trig2.png 
Views:	3 
Size:	4.4 KB 
ID:	25187
    Frankly, I really do not follow your question.
    The above work is all correct.
    Clearly there is no solution: a positive cannot equal a negative.
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  3. #3
    Junior Member Greymalkin's Avatar
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    Re: Derivative algebra

    I understand that, I just want to know how they made the jump from \frac{1}{2}\sqrt{-x}(-1)-1 to \frac{-1}{2\sqrt{-x}}-1
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    Re: Derivative algebra

    Quote Originally Posted by Greymalkin View Post
    I understand that, I just want to know how they made the jump from \frac{1}{2}\sqrt{-x}(-1)-1 to \frac{-1}{2\sqrt{-x}}-1
    The derivative of \sqrt{-x}-x is \frac{-1}{2\sqrt{-x}}-1. They didn't make a jump.
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  5. #5
    Junior Member Greymalkin's Avatar
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    Re: Derivative algebra

    I see that, but I do not understand how that is done algebraically,
    I get \frac{1}{2}\sqrt{-x}(-1)-1

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

    -\frac{\sqrt{-x}}{2}-1
    Which is as far as I get when trying to figure out how it becomes \frac{-1}{2\sqrt{-x}}-1

    Maybe \frac{x}{2\sqrt{-x}}-1 ? closer, but where does the factor of x go?
    Last edited by Greymalkin; October 12th 2012 at 09:12 AM.
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  6. #6
    Junior Member alane1994's Avatar
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    Re: Derivative algebra

    Think of it like this.

    y=\sqrt{f(x)}

    y^{\prime}=\frac{f^{\prime}(x)}{2\sqrt{f(x)}}

    Then for your other parts, differentiate like normal.
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  7. #7
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    Re: Derivative algebra

    Quote Originally Posted by Greymalkin View Post
    I see that, but I do not understand how that is done algebraically,
    I get \frac{1}{2}\sqrt{-x}(-1)-1

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

    -\frac{\sqrt{-x}}{2}-1
    Which is as far as I get when trying to figure out how it becomes \frac{-1}{2\sqrt{-x}}-1

    Maybe \frac{x}{2\sqrt{-x}}-1 ? closer, but where does the factor of x go?
    -\frac{\sqrt{-x}}{2}-1\not=\frac{-1}{2\sqrt{-x}}-1
    It is the case that the are not equal.
    What makes you think they are?
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  8. #8
    Junior Member Greymalkin's Avatar
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    Re: Derivative algebra

    Think I just forgot to use brackets over a root symbol when differentiating, because my work did not include ^(-1/2) which easily works with the above. I always seem to fudge the easy stuff in math.
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  9. #9
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    Re: Derivative algebra

    The problem is NOT the differentiation. I suspect that you did not tell us the whole problem! After differentiating they set the derivative equal to 0 and then started solving for x. That would be a natural thing to do if the problem were not just to find the derivative but to find maximum and minimum values for the function.
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