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Math Help - FTC to find the derivative of this function

  1. #1
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    FTC to find the derivative of this function

    I've attached a picture, I have a good handle on using the fundamental theorem of calculus to find derivatives of functions, but this one is just stumping me for some reason. First I tried doing it as is, then I literally had no idea what I could be doing wrong so I tried negating it so that the sin is on top, but that makes no difference. Could someone help a brotha out with this toughie? The thing that's throwing me is the fact that there isn't a real number on the top or bottom of the integral, they're both trig functions. I didn't think this would affect how to go about solving the problem but it must somehow...
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  2. #2
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    Krizalid's Avatar
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    It's y'={{\left( 5+{{\cos }^{2}}x \right)}^{10}}(\cos x)'-{{\left( 5+{{\sin }^{2}}x \right)}^{10}}(\sin x)'.
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  3. #3
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    Quote Originally Posted by Krizalid View Post
    It's y'={{\left( 5+{{\cos }^{2}}x \right)}^{10}}(\cos x)'-{{\left( 5+{{\sin }^{2}}x \right)}^{10}}(\sin x)'.
    So basically when you have an integral with two variables, one in each value top and bottom, you subtract the bottom from the top, evaluating it for both, correct?
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  4. #4
    Member Abu-Khalil's Avatar
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    If f(x) is a continuos function in (g(x),h(x)) then F(x)=\int_{g(x)}^{h(x)}f(x)dx\Rightarrow F'(x)=f(h(x))h'(x)-f(g(x))g'(x).

    So y'=-(5+\cos^2 x)^{10}\sin x-(5+\sin^2 x)^{10}\cos x.
    Last edited by Abu-Khalil; February 6th 2009 at 01:49 PM. Reason: Read mistake.
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  5. #5
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    Gotcha thanks, I wish my professor would cover the finer details of things like this BEFORE assigning us homework involving them
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  6. #6
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    But there're some typos there, it should be <br />
F(x)=\int_{g(x)}^{h(x)}f(t)\,dt\implies F'(x)=f(h(x))h'(x)-f(g(x))g'(x)<br />
and in one hand it's h(x)=\cos x not \cos^2x.
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  7. #7
    Member Abu-Khalil's Avatar
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    Quote Originally Posted by Krizalid View Post
    But there're some typos there, it should be <br />
F(x)=\int_{g(x)}^{h(x)}f(t)\,dt\implies F'(x)=f(h(x))h'(x)-f(g(x))g'(x)<br />
and in one hand it's h(x)=\cos x not \cos^2x.
    Ups, bad reading. Edited.
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