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Math Help - need help understanding d/dx(x^(1/2))

  1. #1
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    need help understanding d/dx(x^(1/2))

    Hi,
    I'm having problems understanding the derivation of fraction exponents like this for example:
    d/dx(x^(1/2))

    Obviously I have more complicated problems than this to evaluate, but a basic explanation would be helpful.
    The book I'm using for some reason skips talking about this....

    The example I'm looking at on my homework is this:
    d/dx((4/z^(1/3))-(z^(1/4)))

    4 over the 3rd root of z, minus z to the 4th root.

    Thanks much,
    sig
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    Re: need help understanding d/dx(x^(1/2))

    Quote Originally Posted by sig View Post
    Hi,
    I'm having problems understanding the derivation of fraction exponents like this for example:
    d/dx(x^(1/2))
    There is one basic power rule: \frac{d}{dx}\left(x^{\alpha}\right)=\alpha x^{\alpha-1}

    Thus \frac{d}{dx}\left(x^{\frac{1}{2}}\right)=\frac{1}{  2} x^{-\frac{1}{2}}
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    Re: need help understanding d/dx(x^(1/2))

    Quote Originally Posted by Plato View Post
    There is one basic power rule: \frac{d}{dx}\left(x^{\alpha}\right)=\alpha x^{\alpha-1}

    Thus \frac{d}{dx}\left(x^{\frac{1}{2}}\right)=\frac{1}{  2} x^{-\frac{1}{2}}
    I must be making mistakes when I evaluate then, because I know the power rule.
    For instance my answer sheet says

    \frac{d}{dx}\left(x^{\frac{1}{2}}\right)=-\frac{1}{2x^{\frac{1}{2}}}

    I would think the x^(-1/2) would go on top of the fraction, and I don't know how the negative is pulled out of the exponent into the front of the answer.
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    Re: need help understanding d/dx(x^(1/2))

    Quote Originally Posted by sig View Post
    my answer sheet says

    \frac{d}{dx}\left(x^{\frac{1}{2}}\right)=-\frac{1}{2x^{\frac{1}{2}}}

    I would think the x^(-1/2) would go on top of the fraction, and I don't know how the negative is pulled out of the exponent into the front of the answer.
    Your answer sheet is wrong,

    It should be \frac{d}{dx}\left(x^{\frac{1}{2}}\right)=\frac{1}{  2x^{\frac{1}{2}}}

    Recall that \left(x^{-\frac{1}{2}}\right)=\frac{1}{x^{\frac{1}{2}}}
    Last edited by Plato; May 20th 2012 at 01:31 PM.
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    Re: need help understanding d/dx(x^(1/2))

    Quote Originally Posted by Plato View Post
    Your answer sheet is wrong,

    It should be \frac{d}{dx}\left(x^{\frac{1}{2}}\right)=\frac{1}{  2x^{\frac{1}{2}}}

    Recall that \left(x^{-\frac{1}{2}}\right)=\frac{1}{2x^{\frac{1}{2}}}
    Oh I'm sorry, the negative was a part of a different problem.
    Oh, and now I see how the numbers end up on bottom; the negative exponent.

    OK thank you very much, I'll try some equations again and see if I can understand better.

    -sig
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    Re: need help understanding d/dx(x^(1/2))

    Hello, I'm back.
    After looking and working on this equation,

    \frac{d}{dz}\left((4/z^{\frac{1}{3}}\right)-(z^\frac{1}{4}))

    I've come to the pre-simplified answer of:

    -(4/3z^(4/3))-(1/(4z^(3/4))

    And when I look the answer up on WRA, it gives the simplified answer of:

    -((3z^(7/12)+16)/(12z^(4/3))

    and I'm not seeing how this simplification is working.

    Can anyone help explain?

    Thanks..
    -sig
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    Re: need help understanding d/dx(x^(1/2))

    Quote Originally Posted by sig View Post
    Hello, I'm back.
    After looking and working on this equation,

    \frac{d}{dz}\left(4/z^{\frac{1}{3}}\right)
    That is \frac{d}{dz}\left(4z^{-\frac{1}{3}}\right)=\frac{-4}{3}z^{\frac{- 4}{3}}
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    Re: need help understanding d/dx(x^(1/2))

    Quote Originally Posted by Plato View Post
    That is \frac{d}{dz}\left(4z^{-\frac{1}{3}}\right)=\frac{-4}{3}z^{\frac{- 4}{3}}
    Yes I got that much.
    which equals

    -(4/3z^(4/3))

    But I don't see how this:

    -(4/3z^(4/3)) - (1/4z^(/34))

    turns into this:

    - ((3z^(7/12)+16)/(12z^(4/3)))


    Sorry for not entering this into special text- is there a way to convert without manually coding it?
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    Re: need help understanding d/dx(x^(1/2))

    Quote Originally Posted by sig View Post
    Hello, I'm back.
    After looking and working on this equation,

    \frac{d}{dz}\left((4/z^{\frac{1}{3}}\right)-(z^\frac{1}{4}))
    Those parentheses are a little confusing. Do you mean \frac d{dz}\left[\frac4{z^{1/3}}-z^{1/4}\right]?

    We have

    \frac d{dz}\left[\frac4{z^{1/3}}-z^{1/4}\right]

    =\frac d{dz}\left[4z^{-1/3}}-z^{1/4}\right]

    =-\frac43z^{-4/3}-\frac14z^{-3/4}

    =-\frac4{3z^{4/3}}-\frac1{4z^{3/4}}

    =-\frac4{3z^{16/12}}-\frac1{4z^{9/12}}

    and, combining fractions,

    =-\frac{4(4)}{\left(3z^{16/12}\right)(4)}-\frac{3z^{7/12}}{\left(4z^{9/12}\right)\left(3z^{7/12}\right)}

    =\frac{-16-3z^{7/12}}{12z^{4/3}}

    =-\frac{3z^{7/12}+16}{12z^{4/3}}
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    Re: need help understanding d/dx(x^(1/2))

    =-\frac4{3z^{16/12}}-\frac1{4z^{9/12}}
    Yes that's right and this above is where I can get to. But here below I don't understand


    and, combining fractions,

    =-\frac{4(4)}{\left(3z^{16/12}\right)(4)}-\frac{3z^{7/12}}{\left(4z^{9/12}\right)\left(3z^{7/12}\right)}

    =\frac{-16-3z^{7/12}}{12z^{4/3}}

    =-\frac{3z^{7/12}+16}{12z^{4/3}}
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    Re: need help understanding d/dx(x^(1/2))

    nevermind figured it out, thanks!

    ~sig
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    Re: need help understanding d/dx(x^(1/2))

    Quote Originally Posted by sig View Post
    Yes that's right and this above is where I can get to. But here below I don't understand
    To add fractions together we need a common denominator. The least common multiple of 3 and 4 is 12, and we can treat z^{4/3} as a common multiple of z^{4/3} and z^{3/4} because \frac{z^{4/3}}{z^{3/4}} =z^{7/12} \Rightarrow z^{3/4}\cdot z^{7/12} =z^{4/3}. So we shall make our common denominator 12z^{4/3}.

    Now, how do we get this denominator in both fractions? By multiplying the numerator and denominator of the left fraction by 4 (because 4\cdot3z^{4/3}=12z^{4/3}) and by multiplying the numerator and denominator of the right fraction by 3z^{7/12} (because, again, 4z^{3/4}\cdot3z^{7/12} =12z^{4/3}).

    Quote Originally Posted by sig View Post
    nevermind figured it out, thanks!
    I just finished typing this response as you posted, so I'll submit it anyway.
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    Re: need help understanding d/dx(x^(1/2))

    Thanks Reck,

    I had been thinking combining fractional exponents were like combining fractions but I was totally wrong!

    More than helpful, I'll be back sometime soon, I know it.

    ~sig
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