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Thread: Help with function u of x (chain rule)

  1. #1
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    Help with function u of x (chain rule)

    $\displaystyle Given f(x) =\frac{(x^2 +3x +1)^5}{(x+3)^5}$, identify a function of u of x and an integer n not equal to 1 such that $\displaystyle f(x)=u^n$. Then compute $\displaystyle f'(x)$.
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  2. #2
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    Quote Originally Posted by Yogi_Bear_79
    $\displaystyle Given f(x) =\frac{(x^2 +3x +1)^5}{(x+3)^5}$, identify a function of u of x and an integer n not equal to 1 such that $\displaystyle f(x)=u^n$. Then compute $\displaystyle f'(x)$.
    You can redo this to,
    $\displaystyle \left( \frac{x^2+3x+1}{x+3} \right)^5$
    Therefore,
    $\displaystyle u=\frac{x^2+3x+1}{x+3}$
    Thus,
    $\displaystyle \frac{du}{dx}=\frac{(x^2+3x+1)'(x+3)-(x^2+3x+1)(x+3)'}{(x+3)^2}$
    Thus,
    $\displaystyle \frac{du}{dx}=\frac{(2x+3)(x+3)-(x^2+3x+1)(1)}{(x+3)^2}$
    Thus,
    $\displaystyle \frac{du}{dx}=\frac{2x^2+9x+9-x^2-3x-1}{(x+3)^2}$
    Thus,
    $\displaystyle \frac{du}{dx}=\frac{x^2+6x+8}{x^2+6x+9}=\frac{x^2+ 6x+9-1}{x^2+6x+9}=1-(x+3)^{-2}$
    And,
    $\displaystyle \frac{dy}{du}=5u^4$
    Thus,
    $\displaystyle \frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}=5\cdot \left( 1-(x+3)^{-2} \right) \cdot \frac{(x^2+3x+1)^4}{(x+3)^4} $
    Thus,
    $\displaystyle \frac{5(x^2+3x+1)}{(x+3)^4}-5(x^2+3x+1)(x+3)^2$
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