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Graph the functions problem

  1. Jan 12, 2005 #1

    kreil

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    If [tex]y=sin^3(1-2x)[/tex] then [tex]\frac{dy}{dx}=?[/tex]
    a)[tex]3sin^2(1-2x)[/tex]
    b)[tex]-2cos^3(1-2x)[/tex]
    c)[tex]-6sin^2(1-2x)[/tex]
    d)[tex]-6sin^2(1-2x)cos(1-2x)[/tex]
    e)[tex]-6cos^2(1-2x)[/tex]

    Here is my work:

    [tex]\frac{dy}{dx}=3cos^2(1-2x)(-2)=-6cos^2(1-2x)[/tex]

    When I checked the problem with my friend, he had (B) and after we discussed it we decided to graph the functions to find the answer and according to that the answer is (D). Can someone show me the work for that answer, I don't really understand how it could be obtained.
     
  2. jcsd
  3. Jan 12, 2005 #2

    Curious3141

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    Answer is d). Apply the chain rule more carefully. There are three nested functions to consider here : sin(z), y^3, (1 - 2x). You missed out on differentiating the sin(z) part
     
    Last edited: Jan 12, 2005
  4. Jan 12, 2005 #3

    arildno

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    Dearly Missed

    D) is certainly the correct answer.
    When using the chain rule, it is crucial to be clear on what is the outer function and what is the kernel(s).

    In order to develop a proficiency in this, you should begin with being VERY careful
    (as you get more experience, you can drop a few of the intermediate steps:
    Here's one way
    1. Set u(x)=1-2x
    2) Set v(u)=sin(u)
    3) Set w(v)=v^3

    Hence,
    [tex]y(x)=w(v(u(x)))[/tex]
    Or:
    [tex]\frac{dy}{dx}=\frac{dw}{dv}\frac{dv}{du}\frac{du}{dx}[/tex]
    Now, we have:
    [tex]\frac{dw}{dv}=3v^{2},\frac{dv}{du}=\cos(u), \frac{du}{dx}=-2[/tex]

    Now, assemble your answer.
     
  5. Jan 12, 2005 #4

    Galileo

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    The answer is (D). You have to apply the chain rule twice.

    A little elaborate way to use the chain rule is by substitution:
    Let [itex]u=\sin(v)[/itex] and [itex]v=1-2x[/itex]

    Then:
    [tex]y=u^3[/tex]
    and
    [tex]\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}=\frac{dy}{du}\frac{du}{dv}\frac{dv}{du}[/tex]

    Work it out and you'll get (D).
     
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