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Math Help - Finding third derivative

  1. #1
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    Finding third derivative

    y = (5x - 1)^{\frac{1}{2}}
    So to find the 1st derivative I use the Chain Rule:

    y' = \frac{1}{2}(5x - 1)^{\frac{1}{2} - 1} \cdot  (5 - 0) \\<br />
= \frac{5}{2}(5x - 1)^{-\frac{1}{2}}

    2nd:

    Multiplication rule [f'(x)*g(x) + f(x)*g'(x)] and Chain Rule:

    y'' = \frac{5}{2}(5x - 1)^{-\frac{1}{2}} \\<br />
= 0\cdot (5x - 1)^{-\frac{1}{2}} + \frac{5}{2}\cdot [-\frac{1}{2}(5x - 1)^{-\frac{1}{2} - 1} \cdot (5 - 0)] \\<br />
= \frac{5}{2}\cdot (-\frac{5}{2})(5x - 1)^{-\frac{3}{2}} \\<br />
= -\frac{25}{4}(5x - 1)^{-\frac{3}{2}}

    3rd:

    Multiplication rule [f'(x)*g(x) + f(x)*g'(x)] and Chain Rule:

    y''' = -\frac{25}{4}(5x - 1)^{-\frac{3}{2}} \\<br />
= 0\cdot (5x - 1)^{-\frac{3}{2}} + -\frac{25}{4}\cdot [(-\frac{3}{2}(5x - 1)^{-\frac{3}{2}-1}) \cdot (5 - 0)] \\<br />
= -\frac{25}{4}\cdot (-\frac{15}{2})(5x - 1)^{-\frac{5}{2}} \\<br />
= \frac{375}{8}(5x - 1)^{-\frac{5}{2}} \\<br />
= \frac{375}{8(5x-1)^{\frac{5}{2}}}

    I think I got this one wrong but this is how I worked it out...have I done any step incorrectly here?
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  2. #2
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    Re: Finding third derivative

    don't make it so hard on yourself ... on every iteration of a derivative, the derivative of (5x-1) becomes an additional factor

    y = (5x-1)^{1/2}

    y' = \frac{1}{2}(5x-1)^{-1/2} \cdot 5 = \frac{5}{2}(5x-1)^{-1/2}

    you don't need the product rule ... \frac{d}{dx} \left[k \cdot f(x)\right] = k \cdot f'(x)

    y'' = -\frac{5}{4}(5x-1)^{-3/2} \cdot 5 = -\frac{25}{4}(5x-1)^{-3/2}

    y''' = \frac{75}{8}(5x-1)^{-5/2} \cdot 5 = \frac{375}{8}(5x-1)^{-5/2} = \frac{375}{8(5x-1)^{5/2}}
    Thanks from daigo
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    Re: Finding third derivative

    Sorry, I forgot what the rule is called

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

    Which one is this
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    Re: Finding third derivative

    Quote Originally Posted by skeeter View Post
    you don't need the product rule ... \frac{d}{dx} \left[k \cdot f(x)\right] = k \cdot f'(x)
    Oh wow, the use of this basic rule was very clever. If I had thought of it it would've made this so much easier
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  5. #5
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    Re: Finding third derivative

    Quote Originally Posted by daigo View Post
    Sorry, I forgot what the rule is called

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

    Which one is this
    It is the product rule b/c you are taking the derivative of a product i.e f(x) times g(x)
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