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Derivative of Composite Function

  1. Dec 7, 2011 #1
    1. The problem statement, all variables and given/known data

    \begin{equation}
    f(x)=
    \begin{cases}
    5x+2 &, x \leq 1 \\
    3x^2 &, 1<x<2\\
    4-x &, x\geq 2
    \end{cases}
    \end{equation}

    \begin{equation}
    g(x)=
    \begin{cases}
    \frac{1}{5}(2+3 cos x) &, x <0 \\
    4-sin x &, x \geq 0
    \end{cases}
    \end{equation}

    Find [itex] h = f \circ g [/itex] and then by using chain rule to find the derivative of [itex] f \circ g [/itex] .

    2. Relevant equations

    \begin{equation}
    h(x)=f\circ g =
    \begin{cases}
    4+3 cos x &, x <0 \\
    sin x &, x \geq 0
    \end{cases}
    \end{equation}

    3. The attempt at a solution
    From [itex] h(x) = f \circ g [/itex] above, I directly differentiate and I get the following
    \begin{equation}
    h'(x)=
    \begin{cases}
    -3 sin x &, x < 0 \\
    cos x &, x > 0
    \end{cases}
    \end{equation}

    If chain rule are requested in finding the derivative, I know the formula is [itex] h'(x) = f' (g(x)) \cdot g'(x) [/itex]. Any idea to find the derivative using chain rule?

    Thank you.
     
  2. jcsd
  3. Dec 7, 2011 #2

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    It seems overly pedantic to do it explicitly with the chain rule, but if you must do so, just write h(x) as a two piece formula like this:

    If x ≥ 0, f(x) = 4-x, g(x) = 4 - sin(x) and [itex]h=f\circ g[/itex]
    If x < 0 f(x) = 5x+2, g(x) = (1/5)(2+3cos(x)) and [itex]h=f\circ g[/itex]
    Then do the chain rule on each piece.
     
  4. Dec 7, 2011 #3
    Thank you for the reply. I got the idea from it. This is because my lecturer want us to compare whether the answer obtained from directly differentiate [itex]h(x)[/itex] is the same
    as the answer obtained from the chain rule. :smile:
     
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