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Math Help - First Principal derivatives

  1. #1
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    First Principal derivatives

    Hello, im new here but im doing a calculus course online a grade 12 University course and im stuck on how to do first principals derivatives, and since i have no teacher just learning off the book its tough if you guys dont mind help me solve these questions for a practice so i can start to hand in work.

    a) f(x)=x-2x+1
    b) f(x)=x-3x
    c) f(x)=(√2x)
    d) f(x)=4/x+1

    greatly appreciated
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  2. #2
    Super Member craig's Avatar
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    Quote Originally Posted by pr3m View Post
    Hello, im new here but im doing a calculus course online a grade 12 University course and im stuck on how to do first principals derivatives, and since i have no teacher just learning off the book its tough if you guys dont mind help me solve these questions for a practice so i can start to hand in work.

    a) f(x)=x-2x+1
    b) f(x)=x-3x
    c) f(x)=(√2x)
    d) f(x)=4/x+1

    greatly appreciated
    If you are doing calculus then you should be aware of the basics, that is that the derivative of x^n is nx^{n-1}, where n is a constant.

    For your first one, you have f(x)=x^2-2x+1 , f'(x)=2x-2.

    b)  f(x)=x^3-3x^2 - follow the steps above.

    c)  f(x)=\sqrt{2x}, this can be written as 2x^{\frac{1}{2}}, follow the steps above.

    d) f(x)= \frac{4}{x+1}, this can be written as 4(x+1)^{-1}, follow the steps above.

    If there is anything you are unsure about just ask =)
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  3. #3
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    hey thanks for the response... i know how to find the derivative of somthing to solve for f prime its just with first principals they put it in the other form where its like f(x+h)-f(x) / h =... thats what im having trouble with finding the steps to solve each one and geting it where u go lim h -> 0 or w\e know what i mean?
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  4. #4
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    Are you asking to find the derivatives using the limit definition? (BTW, this should have been posted in the Calculus subforum.)

    Then, using this:
    \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}

    For (a), we get
    \lim_{h \to 0} \frac{[(x + h)^2 - 2(x + h) + 1] - [x^2 - 2x + 1]}{h}

    = \lim_{h \to 0} \frac{x^2 + 2xh + h^2 - 2x - 2h + 1 - x^2 + 2x - 1}{h}

    = \lim_{h \to 0} \frac{2xh + h^2 - 2h}{h}

    = \lim_{h \to 0} (2x + h - 2)

    As h \rightarrow 0,
    \lim_{h \to 0} (2x + h - 2) = 2x - 2.


    01
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  5. #5
    Newbie Deco's Avatar
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    Generally, first derivatives for the ones you listed above, involving powers
    and polynomials; x^n \rightarrow nx^{n-1}

    This is a rather sloppy and shorthand way of showing what has
    been demonstrated above.

    (The \longrightarrow represents the process of differentiating)

    So,

    (1) x^5 \longrightarrow 5x^4

    (2)  x^1 \longrightarrow 1x^0 = 1; x \neq 0

    (3) 4x^3 + 5 = 4x^3 +5x^0 \longrightarrow 12x^2 + 0 = 12x^2
    Last edited by Deco; July 25th 2009 at 07:46 PM.
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  6. #6
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    Quote Originally Posted by pr3m View Post
    ... im stuck on how to do first principals derivatives,
    ...

    c) f(x)=(√2x)

    ...
    I'll do this step by step:

    f'(x)=\lim_{ h \to 0}\dfrac{\sqrt{2(x+h)} - \sqrt{2x}}{h}

    {\color{white}f'(x)}=\lim_{ h \to 0}\dfrac{\sqrt{2(x+h)} - \sqrt{2x}}{h} \cdot \dfrac{\sqrt{2(x+h)} + \sqrt{2x}}{\sqrt{2(x+h)} + \sqrt{2x}}

    {\color{white}f'(x)}=\lim_{ h \to 0}\dfrac{2x+2h-2x}{h \left(\sqrt{2(x+h)} + \sqrt{2x} \right)}

    {\color{white}f'(x)}=\lim_{ h \to 0}\dfrac{2}{\sqrt{2(x+h)} + \sqrt{2x} }

    .................................................. .......... Since \lim_{h \to 0} \sqrt{2(x+h)} = \sqrt{2x} the last limit simplifies to:

    {\color{white}f'(x)}= \dfrac{2}{\sqrt{2x} +  \sqrt{2x} }= \dfrac{2}{2 \cdot  \sqrt{2x} }
    Thus:
    \boxed{f'(x)=\dfrac1{\sqrt{2x}}}
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