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Thread: Differentiating Big Multiple

  1. #1
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    Differentiating Big Multiple

    Hi I need urgent help for this expression as I have no idea how to get its derivative:


    $\displaystyle f(x)=(x^2+1)^100(x^3-7)^2(x-1)(x^{10}+x+3)^{-1}$

    I have to use the ln function but I can only come up to here:

    $\displaystyle \frac{d}{dx}ln(f(x))=100\frac{2x}{x^2+1}+2\frac{3x ^2}{x^3-7}+\frac{1}{x-1}-\frac{10x^9+1}{x^{10}+x+3}$

    Specifically I don't know how to get rid of the natural log function on the left side of the equation. I would appreciate if anyone could tell me how?
    Last edited by Kataangel; Oct 14th 2009 at 08:59 AM.
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  2. #2
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    Quote Originally Posted by Kataangel View Post
    Hi I need urgent help for this expression as I have no idea how to get its derivative:


    $\displaystyle f(x)=(x^2+1)^100(x^3-7)^2(x-1)(x^{10}+x+3)^{-1}$

    I have to use the ln function but I can only come up to here:

    $\displaystyle \frac{d}{dx}ln(f(x))=100\frac{2x}{x^2+1}+2\frac{3x ^2}{x^3-7}+\frac{1}{x-1}-\frac{10x^9+1}{x^{10}+x+3}$

    Specifically I don't know how to get rid of the natural log function on the left side of the equation. I would appreciate if anyone could tell me how?
    $\displaystyle \frac{d}{dx}ln(f(x))= \frac{1}{f(x)}f'(x)$
    $\displaystyle therefore \quad \frac{1}{f(x)}f'(x)==\frac{200x}{x^2+1}+\frac{6x^2 }{x^3-7}+\frac{1}{x-1}-\frac{10x^9+1}{x^{10}+x+3}$
    $\displaystyle therefore \quad f'(x)=f(x)\times (\frac{200x}{x^2+1}+\frac{6x^2}{x^3-7}+\frac{1}{x-1}-\frac{10x^9+1}{x^{10}+x+3})$
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  3. #3
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    Hello, Kataangel!

    Hi, I need urgent help for this expression as I have no idea how to get its derivative.
    . .
    Why do you say that? You obviously know how to use logs and to differentiate!

    $\displaystyle f(x)=(x^2+1)^{100}(x^3-7)^2(x-1)(x^{10}+x+3)^{-1}$

    I have to use the ln function but I can only come up to here:

    $\displaystyle \frac{d}{dx}ln(f(x))=100\frac{2x}{x^2+1}+2\frac{3x ^2}{x^3-7}+\frac{1}{x-1}-\frac{10x^9+1}{x^{10}+x+3}$

    Specifically I don't know how to get rid of the natural log function on the left side of the equation.
    I would appreciate if anyone could tell me how.
    Write it like this: .$\displaystyle {\color{blue}y} \;=\;(x^2+1)^{100}(x^3-7)^2(x-1)(x^{10}+x+3)^{-1}$

    Take logs: .$\displaystyle \ln(y) \;=\;100\ln(x^2+1) + 2\ln(x^3-7) + \ln(x-1) - \ln(x^{10}+x+3) $

    Diffrerentiate: .$\displaystyle \frac{1}{y}\cdot\frac{dy}{dx} \;=\;\frac{200x}{x^2+1} + \frac{6x^2}{x^3-7} + \frac{1}{x-1} - \frac{10x^9 + 1}{x^{10}+x+3} $

    Then: .$\displaystyle \frac{dy}{dx} \;=\;y\bigg[\frac{200x}{x^2+1} + \frac{6x^2}{x^3-7} + \frac{1}{x-1} - \frac{10x^9 + 1}{x^{10}+x+3}\bigg] $

    . . . . . $\displaystyle \frac {dy}{dx}\;=\;(x^2+1)^{100}(x^3-7)^2(x-1)(x^{10}+x+3)^{-1}\bigg[\frac{200x}{x^2+1} + \frac{6x^2}{x^3-7} + \frac{1}{x-1} - \frac{10x^9 + 1}{x^{10}+x+3}\bigg] $


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