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Math Help - Need derivative help once more.

  1. #1
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    Need derivative help once more.

    f(x)= x^\frac{6}{9}

    I need help with fractional exponents.

    is it still nx^n-1
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  2. #2
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    Yes , if y=x^n then y' = n x^(n-1).

    So, in your problem , y'=(6/9)(x^((6/9)-1))= (2/3)(x^(-1/3))=2/(3x^(1/3)).
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  3. #3
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    Quote Originally Posted by Zanderist View Post
    f(x)= x^\frac{6}{9}

    I need help with fractional exponents.

    is it still nx^n-1
    first, note that \frac{6}{9} = \frac{2}{3}

    f(x) = x^{\frac{2}{3}}

    f'(x) = \frac{2}{3} x^{\frac{2}{3} - 1} = ?
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  4. #4
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    Quote Originally Posted by skeeter View Post
    first, note that \frac{6}{9} = \frac{2}{3}

    f(x) = x^{\frac{2}{3}}

    f'(x) = \frac{2}{3} x^{\frac{2}{3} - 1} = ?
    f'(x) = \frac{2}{3} x^{-\frac{1}{3}}

    I have another one.

    4\sqrt{t}+\frac{7}{\sqrt{t}}
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  5. #5
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    Quote Originally Posted by Zanderist View Post
    4\sqrt{t}+\frac{7}{\sqrt{t}}
    First rewrite the equation: 4t^{\frac{1}{2}}+7t^{\frac{-1}{2}}
    Now follow your rules nx^{n-1}
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  6. #6
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    Quote Originally Posted by dani View Post
    First rewrite the equation: 4t^{\frac{1}{2}}+7t^{\frac{-1}{2}}
    Now follow your rules nx^{n-1}
    So...this..
    4/2t^{\frac{1}{2}-1}+7/2t^{\frac{-1}{2}-1}

    to this...

    2t^{-\frac{1}{2}}+\frac{7}{2}t^{-\frac{3}{2}}
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  7. #7
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    Okay let's take it up a level...

    \frac{e^x}{(2-1x)}

    Quotient rule I believe.

    I just want to make sure about this

    f(x)=e^x

    It's derivative is...

    f '(x)= e^x \log e
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  8. #8
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    I have a problem I just can't stand.



    I take it to find B you need the second der. for A you just leave it as is.

    I just don't know what is the proper answer.
    Attached Thumbnails Attached Thumbnails Need derivative help once more.-i-loathe-.jpg  
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  9. #9
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    start a new problem with a new thread, otherwise this one can get very confusing.

    thank you.
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