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justinfung
 3 years ago
solving for g'(x) when g(x)=integral from sqrt(x) to x^2, (sin t)(sqrt t) dt
justinfung
 3 years ago
solving for g'(x) when g(x)=integral from sqrt(x) to x^2, (sin t)(sqrt t) dt

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satellite73
 3 years ago
Best ResponseYou've already chosen the best response.1derivative of the integral is the integrand, plus chain rule replace \(t\) by \(x^2\) and multiply by the derivative of \(x^2\) get \[\sin(x^2)\sqrt{x^2}\times 2x\] for the top part bottom part is the same, but make it negative

justinfung
 3 years ago
Best ResponseYou've already chosen the best response.0I think I am suppose to break it up. integral from sqrt x to 0 plus 0 to x^2. But i have no idea why this is the way to do it

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.1yes but you can break up anyway you like

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.1the top part gives what i wrote above, the bottom part is analogous, just make it negative

justinfung
 3 years ago
Best ResponseYou've already chosen the best response.0the bottom can be any number??

justinfung
 3 years ago
Best ResponseYou've already chosen the best response.0I know I'm suppose to use the second fundamental theorem

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.1yes \[\int_{\sqrt{x}}^{x^2}f(t)dt=\int_{\sqrt{x}}^af(t)dt+\int_a^{x^2}f(t)dt\]

justinfung
 3 years ago
Best ResponseYou've already chosen the best response.0My teacher chose 0 though. Is there a reason for that? should I just choose 0 everytime?

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.1take the derivative of the second part, get \(f(x^2)\times 2x\)

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.1the derivative of the first part is \(f(\sqrt{x})\times \frac{1}{2\sqrt{x}}\)

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.1if you like to pick zero, go ahead a pick zero makes no difference

justinfung
 3 years ago
Best ResponseYou've already chosen the best response.0Ok I understand it now. Thanks!!!! :)
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