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Math Help - Derivative using first principles

  1. #1
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    Derivative using first principles

    So I just finished my first week of calc and our weekend homework is to solve this one problem. We have to find the derivative of y= x/ sqrt(3-x) using first principles.
    I have no idea what I'm doing. It seems that every time I try to solve it I end up with a terribly long equation that doesn't seem to simplify. Any useful suggestions are welcome. Just please take into account I have just started learning calc.
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  2. #2
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    Quote Originally Posted by skyblue View Post
    So I just finished my first week of calc and our weekend homework is to solve this one problem. We have to find the derivative of y= x/ sqrt(3-x) using first principles.
    I have no idea what I'm doing. It seems that every time I try to solve it I end up with a terribly long equation that doesn't seem to simplify. Any useful suggestions are welcome. Just please take into account I have just started learning calc.
    You want to find:

    \frac{dy}{dx}=\lim_{h->0}\frac{f(x+h)-f(x)}{h}

    where

    f(x)=\frac{x}{\sqrt{3-x}}

    right?
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  3. #3
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    Quote Originally Posted by adkinsjr View Post
    You want to find:

    \frac{dy}{dx}=\lim_{h->0}\frac{f(x+h)-f(x)}{h}

    where

    f(x)=\frac{x}{\sqrt{3-x}}

    right?
    Yes exactly. Sorry I didn't know how to do the symbols...
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  4. #4
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    Ok, this one is kind of tricky because there is alot of algebra.

    \lim_{h->0}(\frac{x+h}{h\sqrt{3-(x+h)}}-\frac{x}{h\sqrt{3-x}})

    = \lim_{h->0}\frac{h(x+h)\sqrt{3-x}-xh\sqrt{3-(x+h)}}{h\sqrt{3-(x+h)}\sqrt{3-x}}

    = \lim_{h->0}\frac{xh\sqrt{3-x}+h^2\sqrt{3-x}-xh\sqrt{3-(x+h)}}{h\sqrt{3-(x+h)}\sqrt{3-x}}


    Hold on, I'll finish this up for you in a min, I need to restart my computer. sorry, I was having trouble typing the fractions. There's alot of algebra here lol. Do you get this far when you try to work it out?
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  5. #5
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    Hello, skyblue!

    This is the hardest First Principle problem I've ever seen!
    In case adkinsjr doesn't get back soon, I'll crank it out . . .


    Find the derivative of f(x)\:=\: \frac{x}{\sqrt{3-x}} using first principles.

    I recommend breaking up the definition into four steps: . f'(x) \;=\;\lim_{h\to0}\frac{f(x+h) - f(x)}{h}

    . . [1] Find f(x+h) . . . Replace x with x+h ... and simplify.

    . . [2] Subtract f(x) . . . Subtract the original function ... and simplify.

    . . [3] Divide by h . . . Factor and reduce.

    . . [4] Take the limit as h\to0


    Are you ready? . Here we go . . .



    [1] f(x+h) \;=\;\frac{x+h}{\sqrt{3-(x+h)}} \;=\;\frac{x+h}{\sqrt{3-x-h}}



    [2] f(x+h) - f(x) \;=\;\frac{x+h}{\sqrt{3-x-h}} - \frac{x}{\sqrt{3-x}}  \;=\;\frac{(x+h)\sqrt{3-x} - x\sqrt{3-x-h}}{\sqrt{3-x}\cdot\sqrt{3-x-h}}


    . . Multiply top and bottom by the conjugate of the numerator:

    . . . . \frac{(x+h)\sqrt{3-x} - x\sqrt{3-x-h}}{\sqrt{3-x}\cdot\sqrt{3-x-h}}\cdot\frac{(x+h)\sqrt{3-x} + x\sqrt{3-x-h}}{(x+h)\sqrt{3-x} + x\sqrt{3-x-h}} \;= .  \frac{(x+h)^2(3-x) - x^2(3-x-h)}{\sqrt{3-x}\cdot\sqrt{3-x-h}\cdot[(x+h)\sqrt{3=x} + x\sqrt{3-x-h}]}

    . . which simplifies to: . \frac{6xh - x^2h - xh^2 + 3h^2}{\text{denominator}} \;=\; \frac{h\,(6x - x^2 - xh + 3h)}{\text{denominator}}



    [3] \frac{f(x+h) - f(x)}{h} \;=\;\frac{1}{{\color{red}\rlap{/}}h}\cdot\frac{{\color{red}\rlap{/}}h\,(6x - x^2 - xh + 3h)}{\text{denominator}} \;=\;\frac{6x - x^2 - xh + 3h}{\text{denominator}}



    [4] f'(x) \;=\;\lim_{h\to0}\left[ \frac{6x-x^2 - xh + 3h}{\sqrt{3-x}\cdot\sqrt{3-x-h}\cdot[(x+h)\sqrt{3-x} + x\sqrt{3-x-h}]}\right]

    . . . . . . = \;\frac{6x - x^2 - 0 + 0}{\sqrt{3-x}\cdot\sqrt{3-x-0}\cdot\left[(x+0)\sqrt{3-x} + x\sqrt{3-x-0}\right]}

    . . . . . . = \;\frac{6x-x^2}{\sqrt{3-x}\cdot\sqrt{3-x}\cdot(x\sqrt{3-x} + x\sqrt{3-x})}

    . . . . . . = \;\frac{6x - x^2}{(3-x)2x\sqrt{3-x}} \;\;=\;\;\frac{6x-x^2}{2x(3-x)^{\frac{3}{2}}} \;\;=\;\;\frac{{\color{red}\rlap{/}}x(6-x)}{2{\color{red}\rlap{/}}x(3-x)^{\frac{3}{2}}} \;\;=\;\; \boxed{\frac{6-x}{2(3-x)^{\frac{3}{2}}}}


    I need a nap!
    .
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  6. #6
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    Wow...I was watching that and it is a tough one to assign. Significantly easier with the quotient rule...but that wasn't part of the directions...
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  7. #7
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    Thanks folks, I started to have trouble with that one too. Students are usually asked to find the derivatives of functions like y=x^2 or y=\sqrt{1-x} using this method. More complex derivatives are just found using formulae for products and quotients, which can be applied with zero intelligence.
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