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Thread: Calculus n-th derivatives

  1. #1
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    Calculus n-th derivatives

    Find a formula of f^(n)(x), if f(x)=(4+x)^(-1).

    Find the third degree polynomial Q such that Q(1)=-1, Q'(1)=1, Q''(1)=8, and Q'''(1)=18.

    Suppose that f(x)=g(x)/x^3 and that g is twice differentiable. Find f'' in terms of g, g', and g''.

    Thanks!
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  2. #2
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    Quote Originally Posted by asnxbbyx113 View Post

    Suppose that f(x)=g(x)/x^3 and that g is twice differentiable. Find f'' in terms of g, g', and g''.
    That,
    $\displaystyle f'=\frac{g'x^3-3x^2g'}{x^6}$
    $\displaystyle f'=\frac{g'}{x^3}-\frac{3g'}{x^4}$
    $\displaystyle f''=\frac{g''x^3-3x^2g'}{x^6}-\frac{3g''x^4-12x^3g'}{x^8}$.
    But I am not going to simplify that.
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    Hello, asnxbbyx113!

    Find a formula for $\displaystyle f^{(n)}(x)$, if $\displaystyle f(x) = (4+x)^{-1}$

    Start cranking out derivatives and look for a pattern . . .

    $\displaystyle \begin{array}{cccccc}f(x) & = & (4 + x)^{-1} \\
    f'(x) & = & \text{-}1(4 + x)^{-2} \\
    f''(x) & = & 1\!\cdot\!2(4 + x)^{-3} \\
    f'''(x) & = & \text{-}1\!\cdot\!2\!\cdot\!3(4 + x)^{-4} \\
    f^{(4)}(x) & = & 1\!\cdot\!2\!\cdot\!3\!\cdot\!4(4 + x)^{-5} \\
    \vdots & & \vdots \end{array}$

    Therefore: .$\displaystyle f^{(n)}(x)\;=\;(\text{-}1)^n\,n!\,(4 + x)^{\text{-}(n+1)} $

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  4. #4
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    Hello again, asnxbbyx113!

    Find the third-degree polynomial $\displaystyle Q(x)$ such that:
    . . $\displaystyle Q(1)=-1,\;\;Q'(1)=1,\;\; Q''(1)=8,\;\;Q'''(1)=18$

    Let $\displaystyle q(x) \;=\;ax^3 + bx^2 + cx + d$

    Then: .$\displaystyle \begin{array}{cccc}Q(x) & = & ax^3 + bx^2 + cx + d \\ Q'(x) & = & 3ax^2 + 2bx + c \\ Q''(x) & = & 6ax + 2b \\ Q'''(x) & = & 6a\end{array}$


    Since $\displaystyle Q'''(1) = 18\!:\;\;6a \:=\:18\quad\Rightarrow\quad a \,=\,3$

    Since $\displaystyle Q''(1) = 8\!:\;\;6(3)(1) + 2b\:=\:8\quad\Rightarrow\quad b \,=\,\text{-}5$

    Since $\displaystyle Q'(1) = 1\!:\;\;3(3)(1^2) + 2(\text{-}5)(1) + c \:=\:1\quad\Rightarrow\quad c\,=\,2$

    Since $\displaystyle Q(1) = \text{-}1\!:\;\;(3)(1^3) + (\text{-}5)(1^2) + (2)(1) + d \:=\:\text{-}1\quad\Rightarrow\quad d\,=\,\text{-}1$


    Therefore: .$\displaystyle Q(x) \:=\:3x^3 - 5x^2 + 2x - 1$

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