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Analysis Problem (finding a derivative)

  1. Oct 26, 2004 #1

    AKG

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    Suppose that h : [itex]\mathbb{R} \rightarrow \mathbb{R}[/itex] is continuous.
    Calculate f', if f : [itex]\mathbb{R}^2 \rightarrow \mathbb{R}[/itex] is the function:

    [tex]f(x, y) = \int _{\sin (xy)} ^{\cos (xy)} h(t)dt[/tex]

    I have:

    [tex]f' = \left [D_1f(x, y) \ \ \ \ D_2f(x, y)\right ] = \left [\frac{\partial f}{\partial x} \ \ \ \ \frac{\partial f}{\partial y}\right ][/tex]

    [tex] = \left [\frac{\partial}{\partial x}\int_{\sin (xy)} ^{\cos (xy)} h(t)dt \ \ \ \ \frac{\partial}{\partial y}\int_{\sin (xy)} ^{\cos (xy)} h(t)dt\right ][/tex]

    Let H be the antiderivative of h (can I assume it exists from the continuity of h? I would think so), then:

    [tex]f' = \left [\frac{\partial}{\partial x}H(\cos xy) - H(\sin xy) \ \ \ \ \frac{\partial}{\partial y}H(\cos xy) - H(\sin xy)\right ][/tex]

    [tex]f' = -\left [y(\sin (xy)h(\cos (xy)) + \cos (xy)h(\sin (xy))) \ \ \ \ x(\sin (xy)h(\cos (xy)) + \cos (xy)h(\sin (xy)))\right ][/tex]

    Is this right? Thanks.
     
    Last edited: Oct 26, 2004
  2. jcsd
  3. Oct 26, 2004 #2

    HallsofIvy

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    Yes.
    (and other stuff to make more than 10 characters)
     
  4. Oct 26, 2004 #3

    AKG

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    Thanks......
     
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