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Callisto
 3 years ago
Find \(\frac{df}{dx}\) where \[f(x) = \int_{x}^{x^2}\frac{1}{\sqrt{1+t^6}}dt\] for x>0 How to start??
Callisto
 3 years ago
Find \(\frac{df}{dx}\) where \[f(x) = \int_{x}^{x^2}\frac{1}{\sqrt{1+t^6}}dt\] for x>0 How to start??

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zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.2So this is one of those problems that involves requires you to apply the FTC Part 1. The Fundamental Theorem of Calculus part 1 is,\[\huge \frac{d}{dx}\int\limits_c^x f(t) dt=f(x)\] With this problem, we have an function of x within the upper and lower limits, so it'll look a bit different.\[\large \frac{d}{dx}\int\limits_{x}^{x^2}f(t)dt=f(x^2)\color{blue}{(\frac{d}{dx}x^2)}f(x)\color{blue}{(\frac{d}{dx}x)}\]The blue pieces are there because we have to apply the chain rule.

tkhunny
 3 years ago
Best ResponseYou've already chosen the best response.1Fundamental Theorem of Integral Calculus will get you started. \(f(x) = \int\limits_{x}^{x^{2}}g(t)\;dt\;=\;G(x^{2})  G(x)\) Where \(G(x)\) is an antiderivative of \(g(x)\). Chain Rule will get us the rest of the way. \(\dfrac{d}{dx}\left(G(x^{2})  G(x)\right)\;=\;g(x^{2})\cdot (2x)  g(x)\cdot (1)\)

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.0I'm sorry... I was so blind... Thanks all!!!!

tkhunny
 3 years ago
Best ResponseYou've already chosen the best response.1Don't forget that we DO have to believe that the antiderivative EXISTS! Good work.

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.0f(x) = f(x) too, I guess. Since it's an even function..
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