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Help chain rule/tangent stuff

  1. Dec 13, 2004 #1
  2. jcsd
  3. Dec 13, 2004 #2
    chain rule states that if [itex]f(x)=f(u(x))[/itex] then

    [tex]\frac{df}{dx}=\frac{df}{du}\frac{du}{dx}[/tex]

    as examples, first one

    [tex]f(x)=\frac{7}{(4x^3-6x^2)^3}[/tex]

    [tex]u(x)=4x^3-6x^2[/tex]

    [tex]f(u)=\frac{7}{u^3}[/tex]

    given that

    [tex]\frac{df}{du}=-\frac{21}{u^4}[/tex]

    and

    [tex]\frac{du}{dx}=12x^2-12x[/tex]

    then chain rule states

    [tex]\frac{df}{dx}=\frac{df}{du}\frac{du}{dx}=-\frac{21}{u^4}(12x^2-12x)[/tex]

    substituting [itex]u[/itex]

    [tex]\frac{df}{dx}=-\frac{252}{(4x^3-6x^2)^4}(x^2-x)[/tex]

    in the next one, you are almost there... all you have to do is factorize [itex](6x^2-5)^4(2x-1)^3[/itex]

    the third one you can do it by calculating [itex]y=y(x)[/itex] and [itex]x=x(y)[/itex] and then differentiate ie

    [tex]y(x)=\frac{x}{4}(3 \pm \sqrt{5})[/tex]

    or by implicit differentiation.

    The next one you already have it, the only thing you need to do is evaluate [itex]y'(2)[/itex] and [itex]y'(3)[/itex] then use the fact that
    [tex]m=\frac{f'(3)-f'(2)}{3-2}[/tex]

    in the last two you use the fact that [itex](f\circ g)(x)=g(f(x))[/itex] be carefull tough, you are doing it wrong in the last one (watch the root).
     
    Last edited: Dec 13, 2004
  4. Dec 14, 2004 #3
    Thaaaaank you

    Thaaaaank you :biggrin:
     
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