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Thread: Derivative of composite function

  1. #1
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    Derivative of composite function



    so we have f(g(x))

    U=g(x)
    f'(U)*U*U'
    when, x=1, U=3 & U'=2
    f'(3)*3*2
    -4*3*2=-12

    -12 is totally wrong apparently. Can someone please explain how its done?
    Thanks!
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  2. #2
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    If this is graded homework, then it is our policy not to give an answer.
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  3. #3
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    its not graded, I have unlimited attempts.
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  4. #4
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    Do you know the "chain rule"? That is what is being tested here. Using your notation, with U= g(x), f(g(x))= f(U) and the derivative is f'(U)U', NOT the "f'(U)UU'" that you have.
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  5. #5
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    Hello, Vamz!

    $\displaystyle \text{Given the following table of values, find }(f\circ g)'(1)$

    . . $\displaystyle \begin{array}{|c||c|c|c|c|}
    x & f(x) & g(x) & f'(x) & g'(x) \\ \hline \hline
    1 & 2 & 3 & 2 & 4 \\ \hline
    2 & 3 & 1 & -1 & -2 \\ \hline
    3 & 1 & 2 & 4 & 2 \\ \hline\end{array}$

    We have: .$\displaystyle f\circ g \;=\;f(g(x)) $

    Then: .$\displaystyle (f\circ g)' \;=\;f'(g(x))\cdot g'(x)$

    And: .$\displaystyle (f\circ g)'(1) \;=\;f'(g(1))\cdot g'(1)$
    . . . . . . . . . . . . . . . . . $\displaystyle \uparrow\qquad\;\; \uparrow $
    . - - - . . . . . . . . . . . . .$\displaystyle 3 \qquad\;\;4$

    . . . . . . . . . . . . $\displaystyle =\;\;\;f'(3)\cdot 4 $

    . . . . . . . . . . . . $\displaystyle =\quad\;\;4\cdot 4 $

    . . . . . . . . . . . . $\displaystyle =\qquad 16 $

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