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

Mathematics
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So 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.
Fundamental 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. \(\dfrac{d}{dx}\left(G(x^{2}) - G(-x)\right)\;=\;g(x^{2})\cdot (2x) - g(-x)\cdot (-1)\)

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I'm sorry... I was so blind... Thanks all!!!!
Don't forget that we DO have to believe that the antiderivative EXISTS! Good work.
f(-x) = f(x) too, I guess. Since it's an even function..

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