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Math Help - Help with Differentiation

  1. #1
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    Help with Differentiation

    How would I go about doing a problem like this?

    Find dy/dx for y= x^(9/8)

    Just one more thing, where can I find a list to all the commands for MATH tags?
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Zanderist View Post
    How would I go about doing a problem like this?

    Find dy/dx for y= x^(9/8)
    If y = x^n, where n is a constant, then \frac {dy}{dx} = nx^{n - 1}


    Just one more thing, where can I find a list to all the commands for MATH tags?
    see the sticky threads here
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  3. #3
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    Quote Originally Posted by Jhevon View Post
    If y = x^n, where n is a constant, then \frac {dy}{dx} = nx^{n - 1}

    My problem as it was x^\frac{9}{8}

    \frac {dy}{dx} = \frac {9x^\frac{9}{8}-1}{8}

    The "1" by the way is with the Exponent.

    So with that said it should come out to be:

    \frac {9x^8}{8}
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  4. #4
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    Quote Originally Posted by Zanderist View Post
    \frac {dy}{dx} = \frac {9x^\frac{9}{8}-1}{8}

    The "1" by the way is with the Exponent.

    So with that said it should come out to be:

    \frac {9x^8}{8}
    \frac{9}{8}-1 = \frac{1}{8}
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  5. #5
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    y= x^(9/8)
    iff
    ln(y)=ln(x^(9/8))
    ln(y)=(9/8)ln(x), property of logs.
    (dy/dx)(1/y)=(9/8)(1/x)
    (dy/dx)=y(9/8)(1/x)
    (dy/dx)=(x^(9/8))(9/8)(1/x)
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  6. #6
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    Quote Originally Posted by Chris11 View Post
    y= x^(9/8)
    iff
    ln(y)=ln(x^(9/8))
    ln(y)=(9/8)ln(x), property of logs.
    (dy/dx)(1/y)=(9/8)(1/x)
    (dy/dx)=y(9/8)(1/x)
    (dy/dx)=(x^(9/8))(9/8)(1/x)
    and x^(9/8)(1/X)= x^(1/8) as before.

    Do you always use logarithms to differentiate x^n?
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