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Thread: chain rule second derivative

  1. #1
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    chain rule second derivative

    I'm supposed to find the second derivative of $\displaystyle (x^2+5)^8$
    What I got was: $\displaystyle f(u)=u^8, u(x)=(x^2+5)$
    $\displaystyle (8(x^2+5)^7)(2)=16(x^2+5)^7$
    $\displaystyle f(u)=16(u)^7, u(x)=(x^2+5)$
    $\displaystyle (16)7(x^2+5)^6(2)=224(x^2+5)^6$
    but the back of the book says it's: $\displaystyle 80(x^2+5)^6(3x^2+1)$
    Any help would be much appreciated!
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  2. #2
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    $\displaystyle f = (x^2 + 5)^8$.

    Let $\displaystyle u = x^2 + 5$ so that $\displaystyle f = u^8$.


    $\displaystyle \frac{du}{dx} = 2x$.

    $\displaystyle \frac{df}{du} = 8u^7 = 8(x^2 + 5)^7$.


    Therefore $\displaystyle \frac{df}{dx} = 16x(x^2 + 5)^7$.


    Now you will need to use the product rule...

    $\displaystyle \frac{d^2f}{dx^2} = 16x\cdot 2x \cdot 7(x^2 + 5)^6 + 16(x^2 + 5)^7$

    $\displaystyle = 224x^2(x^2 + 5)^6 + 16(x^2 + 5)^7$

    $\displaystyle = 16(x^2 + 5)^6(14x^2 + x^2 + 5)$

    $\displaystyle = 16(x^2 + 5)^6(15x^2 + 5)$

    $\displaystyle = 80(x^2 + 5)^6(3x^2 + 1)$.
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  3. #3
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    oh duh, 2x not 2. thanks!
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