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Math Help - Differentiating

  1. #1
    Member VitaX's Avatar
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    Differentiating

    Differentiate the function and find the slope of the tangent line at the given value of the independent variabale.

    y = (x + 1)^3, x = -2

    \underset{h\to 0}{\mathop{\lim }}\frac{f(x + h) - f(x)}{h}
    \underset{h\to 0}{\mathop{\lim }}\frac{(x + h + 1)^3 - (x + 1)^3}{h}
    = \frac{(1+x^3+3x^2h+3x^2+3h^2x+6hx+3x+h^3+3h^2+3h) - (x^3+3x^2+3x+1)}{h}
    =\frac{h(3x^2+3hx+6x+h^2+3h+3)}{h}
    =3x^2 + 3hx + 6x + h^2 + 3h + 3
    =3x^2 + 6x + 3
    y'=3x^2 + 6x + 3 \rightarrow y'=3(-2)^2 + 6(-2) + 3 \rightarrow y'=3

    Correct?
    Last edited by VitaX; September 23rd 2009 at 11:24 PM.
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  2. #2
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    mr fantastic's Avatar
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    Quote Originally Posted by VitaX View Post
    Differentiate the function and find the slope of the tangent line at the given value of the independent variabale.

    y = (x + 1)^3, x = -2

    \underset{h\to 0}{\mathop{\lim }}\frac{f(x + h) - f(x)}{h}
    \underset{h\to 0}{\mathop{\lim }}\frac{(x + h + 1)^3 - (x + 1)^3}{h}
    = \frac{(1+x^3+3x^2h+3x^2+3h^2x+6hx+3x+h^3+3h^2+3h) - (x^3+3x^2+3x+1)}{h}

    Correct route so far?
    Yes. But is there any particular reason why you getting the derivative from first principles?
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  3. #3
    Member VitaX's Avatar
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    Quote Originally Posted by mr fantastic View Post
    Yes. But is there any particular reason why you getting the derivative from first principles?
    What do you mean exactly first principles? You mean the route I took as opposed to other means? Well I've only learned this method so far.
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