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Math Help - Finding a derivative using first principles

  1. #1
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    Finding a derivative using first principles

    I'm working on this question:

    if f(x) = x^3, then the derivative is 3x^2

    I have to prove this using the definition of the derivative. I got as far as this:

    ((x^3)h + 2(h^2)(x^2)+(h^3)x-x^3)/(h)

    I'm unsure of how to simplify it in order to find the limit as h -> 0.
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  2. #2
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    lim h --> 0 of x^3 = [(x+h)^3 - x^3]/h
    = [(x+h)(x+h)(x+h) - x^3]/h
    = [x^3 + x^2 h + 2 x^2 h + 2 x h^2 + x h^2 + h^3 - x^3]/h
    cancel the x^3 and the h's with the one in the bottom
    = x^2 + 2 x^2 + 2 x h + x h + h^2
    plug in 0 for h
    = x^2 + 2 x^2
    combine like terms
    = 3 x^2

    I hope that helps
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  3. #3
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    Ahh, thanks! The problem was with my simplification of (x+h)^2
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  4. #4
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    I have a similar question:

    Find the derivative of 1/x using first principles. Once again, I'm having trouble simplifying. :/

    I keep getting 0/x^2 when it should be -1/x^2
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  5. #5
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    Quote Originally Posted by Glitch View Post
    I have a similar question:

    Find the derivative of 1/x using first principles. Once again, I'm having trouble simplifying. :/

    I keep getting 0/x^2 when it should be -1/x^2
    Please show all your working. (And in future please post a new question in a new thread).
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  6. #6
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    Working:

    \frac{\frac{1}{x+h} - \frac{1}{x}}{h}

    \frac{h}{x+h} - \frac{h}{x}

    \frac{hx - h(x+h)}{x(x+h)}

    \frac{hx - h(x+h)}{x^2 + hx}

    As h -> 0,

    0/x^2

    Which is incorrect.
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  7. #7
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    Quote Originally Posted by Glitch View Post
    Working:

    \frac{\frac{1}{x+h} - \frac{1}{x}}{h}

    \frac{h}{x+h} - \frac{h}{x} Mr F says: This is wrong. All that follows from it will therefore be wrong. It should be {\color{red}\frac{1}{(x+h)h} - \frac{1}{xh}}.

    \frac{hx - h(x+h)}{x(x+h)}

    \frac{hx - h(x+h)}{x^2 + hx}

    As h -> 0,

    0/x^2

    Which is incorrect.
    ..
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