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Math Help - Derivative of an integral

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    Derivative of an integral

    Just having some trouble with this last problem of mine, its the derivative of the integral (u+5)/(u-8) from 9x to 3x
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    Quote Originally Posted by AlbatrossIV View Post
    Just having some trouble with this last problem of mine, its the derivative of the integral (u+5)/(u-8) from 9x to 3x
    How is u related to x?
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    thats where my confusion comes from, I'm not sure how to incorporate the u's and x's
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  5. #5
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    Quote Originally Posted by AlbatrossIV View Post
    Just having some trouble with this last problem of mine, its the derivative of the integral (u+5)/(u-8) from 9x to 3x
    So the problem is
    \frac{d}{dx}\int_{9x}^{3x} \frac{u+5}{u-8} du?
    You should know that, by the "fundamental theorem of calculus", \frac{d}{dx}\int_a^x f(t)dt= f(x)
    It is a little bit more complicated by the fact that the limits of integration are 9x and 3x but you can do this:
    \int_{9x}^{3x} \frac{u+5}{u-8}du= \int_0^{3x}\frac{u+5}{u-8}du- \int_0^{9x} \frac{u+5}{u-8}du

    And the fact that the upper limit is not just "x" can be handled by setting y= 3x in the first integral and y= 9x in the second:
    The derivative of \int_0^y \frac{u+5}{u-8}du [b]with respect to y[/tex] is just \frac{y+5}{y-8} and the derivative with respect to x is (by the chain rule) that times the derivative of y with respect to x.
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