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Thread: differentiating polynmials

  1. #1
    Senior Member furor celtica's Avatar
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    differentiating polynmials

    the sum of the two shorter sides of a right-angled triangle is 18cm. calculate the least possible length of the hypotenuse.
    ok so obviously i get sqrt(x^2 + (18-x)^2) but how do i differentiate the whole expression from here?
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  2. #2
    MHF Contributor Also sprach Zarathustra's Avatar
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    f(x)=\sqrt{(x^2 + (18-x)^2) }

    f(x)=\sqrt{2x^2-36x+324}

    f'(x)=\frac{4x-36}{2\sqrt{2x^2-36x+324}}=\frac{2x-18}{\sqrt{2x^2-36x+324}}
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  3. #3
    MHF Contributor Unknown008's Avatar
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    Hm... maybe going into more details would be better...

    It's like differentiating (f(x))^n where n \neq 1

    The derivative is:

    n (f(x))^{n-1} \times f'(x)

    So, using your problem;

    f(x) = \sqrt{x^2 + (18-x)^2} = (2x^2 -36x + 324)^{\frac{1}{2}}

    f'(x) = \frac{1}{2} (2x^2 -36x + 324)^{-\frac{1}{2}} \times (4x - 36)

    Which you then simplify to:

    f'(x) = \frac{2x - 18}{\sqrt{2x^2 -36x + 324}}
    Last edited by Unknown008; Aug 16th 2010 at 10:42 AM. Reason: LaTeX typo about sqrt
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  4. #4
    MHF Contributor
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    Quote Originally Posted by furor celtica View Post
    the sum of the two shorter sides of a right-angled triangle is 18cm. calculate the least possible length of the hypotenuse.
    ok so obviously i get sqrt(x^2 + (18-x)^2) but how do i differentiate the whole expression from here?
    note that if you minimize x^2 + (18-x)^2 , you will also minimize its square root.
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  5. #5
    Senior Member furor celtica's Avatar
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    but how does this fit in with f'(x)=0?
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  6. #6
    MHF Contributor Also sprach Zarathustra's Avatar
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    2x^2 -36x + 324 != 0 for all x in R.

    Hence sqrt(2x^2 -36x + 324) !=0 for all x in R.

    ==>

    2x-18=0
    x=9
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  7. #7
    MHF Contributor Unknown008's Avatar
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    Wait! If the denominator becomes 0, the f'(x) becomes infinite!

    It's the numerator which you should set to zero, not the denominator.
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