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Math Help - 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!

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

    . . \begin{array}{|c||c|c|c|c|}<br />
x & f(x) & g(x) & f'(x) & g'(x) \\ \hline \hline<br />
1 & 2 & 3 & 2 & 4 \\ \hline<br />
2 & 3 & 1 & -1 & -2 \\ \hline<br />
3 & 1 & 2 & 4 & 2 \\ \hline\end{array}

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

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

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

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

    . . . . . . . . . . . . =\quad\;\;4\cdot 4

    . . . . . . . . . . . . =\qquad 16

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