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Math Help - Calculus Derivative.

  1. #1
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    Calculus Derivative.

    1) f(x) = (x-3)x^\frac{1}{2} <--- to the one half power.

    2) g(x) = (x^2-3x)^-2 + 2 <---- to the -2 power plus 2

    3) h(x) = (9-x^2)^2


    Thanks for any help.
    Last edited by teddybear; October 25th 2008 at 07:09 PM.
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  2. #2
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    1) f(x) = (x^1-3)x^{\frac{1}{2}}

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

    Now, differentiate, got it ???

    2) g(x) = \frac{1}{(x^2+3x)^2}+2

    g(x) = \frac{1}{x^4+6x^3+9x^2}+2

    Now, differentiate, using Quotient Rule, got it ???

    3) h(x) = (9-x^2)^2

    <br />
h(x) = 81-18x^2+x^4

    Now, differentiate, got it ???
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Shyam View Post

    3) h(x) = (9-x^2)^2

    <br />
h(x) = 81-18x^2+x^4

    Now, differentiate, got it ???
    Or use the chain rule

    \frac{d}{dx}\bigg[f(g(x))\bigg]=f'(g(x))\cdot{g'(x)}
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  4. #4
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    Quote Originally Posted by Mathstud28 View Post
    Or use the chain rule

    \frac{d}{dx}\bigg[f(g(x))\bigg]=f'(g(x))\cdot{g'(x)}

    When I use the chain rule for 3.

    Would it be 2(9-x^2)(-2x)

    = -4x(9-x^2)
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by teddybear View Post
    When I use the chain rule for 3.

    Would it be 2(9-x^2)(-2x)

    = -4x(9-x^2)
    that's correct
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  6. #6
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    I think I finished it... but I'm not sure if I did it right.. Can someone check for me?

    1. {\frac{3}{2}}x^{\frac{1}{2}}-{\frac{3}{2}}x^{\frac{-1}{2}}

    For number 2, I have a problem. After I used the quotient rule, I came upon
    -4x^3-18x^2-18x/(x^4+6x^3+9x^2)^2 and I don't know how to simplify that.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by teddybear View Post
    I think I finished it... but I'm not sure if I did it right.. Can someone check for me?

    1. {\frac{3}{2}}x^{\frac{1}{2}}-{\frac{3}{2}}x^{\frac{-1}{2}}
    yes
    For number 2, I have a problem. After I used the quotient rule, I came upon
    -4x^3-18x^2-18x/(x^4+6x^3+9x^2)^2 and I don't know how to simplify that.
    why use the quotient rule? as the problem was written, it seems to suggest you should use the chain rule. much in the same way you did problem 3
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  8. #8
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    Quote Originally Posted by teddybear View Post
    1) f(x) = (x-3)x^\frac{1}{2} <--- to the one half power.

    2) g(x) = (x^2-3x)^-2 + 2 <---- to the -2 power plus 2

    3) h(x) = (9-x^2)^2


    Thanks for any help.
    Yes, the chain rule is easier for 2).

     g = (x^2 - 3x)^{-2} + 2.

    Let u = x^2 - 3x so  g = u^{-2} + 2.


    \frac{du}{dx} = 2x, \frac{dg}{du} = -2u^{-3} = -2(x^2 - 3x)^{-3}

    So \frac{dg}{dx} = \frac{du}{dx}\times\frac{dg}{du} = -4x(x^2 - 3x)^{-3}.
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