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Thread: Nasty derivative

  1. #1
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    Nasty derivative

    Once again. After missing a week of school and a test, its not fun to catch up, especially in calculus.

    Heres the question.

    Find the slope of the tangent to $\displaystyle h(x)=2x(x+1)^3(x^2+2x+1)^2$ at
    $\displaystyle x=-2$
    Explain how to find the equation of the normal at x= -2.

    Ok so if anyone solves this, can they please do a step by step break down because im about to eat my book.
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  2. #2
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    $\displaystyle
    h(x)=2x(x+1)^3(x^2+2x+1)^2
    $

    $\displaystyle g(x)=2x(x+1)^3 = q(x)p(x) $
    where $\displaystyle q(x) = 2x$
    and $\displaystyle p(x) =(x+1)^3$

    $\displaystyle f(x)=(x^2+2x+1)^2$

    $\displaystyle f'(x) = 2(x^2+2x+1)(2x+2)$

    $\displaystyle g'(x) = q(x)p'(x) + q'(x)p(x) = 2x(3(x+1)^2) + 2(x+1)^3$
    _________________________________________________

    $\displaystyle h(x) = g(x)f(x) \Rightarrow h'(x) = g(x)f'(x) + g'(x)f(x)$

    $\displaystyle h(x) = 2x(x+1)^3 2(x^2+2x+1)(2x+2) + (2x(3(x+1)^2) + 2(x+1)^3)(x^2+2x+1)^2$

    simplify and plug in x=-2! oh and check my work, this thing's a mess!
    Last edited by iknowone; Mar 18th 2008 at 09:56 PM. Reason: x in g'(x) should have been 2x (copy and paste mistake)
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  3. #3
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    eeek. i just worked through it and it gave me -24. the answer in the book says -30. Il double check my math
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  4. #4
    Behold, the power of SARDINES!
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    let

    $\displaystyle h(x)=2x(x+1)^3(x^2+2x+1)^2
    $

    so $\displaystyle h(-2)=1(-2)(-1)(1)^2=4$

    talking the log of both sides we get

    $\displaystyle ln(h(x))=ln(2x)+3ln(x+1)+2ln(x^2+2x+1)$ take the derivative

    $\displaystyle \frac{1}{h}\frac{dh}{dx}=\frac{1}{x}+\frac{3}{x+1} +\frac{2(2x+2)}{x^2+2x+1}$ solving for the derivative...

    $\displaystyle \frac{dh}{dx}=h(x)[\frac{1}{x}+\frac{3}{x+1}+\frac{2(2x+2)}{x^2+2x+1}]$ eval at x=-2

    $\displaystyle \frac{dh}{dx}|_{x=-2}=h(-2)[ \frac{1}{-2}+\frac{3}{-2+1}+\frac{2(2(-2)+2)}{(-2)^2+2(-2)+1}]$

    $\displaystyle \frac{dh}{dx}|_{x=-2}=4[\frac{-1}{2}-3-4]=-30$
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