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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??
 one year ago
 one year 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??
 one year ago
 one year ago

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zepdrixBest ResponseYou've already chosen the best response.2
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.
 one year ago

tkhunnyBest ResponseYou've already chosen the best response.1
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. \(\dfrac{d}{dx}\left(G(x^{2})  G(x)\right)\;=\;g(x^{2})\cdot (2x)  g(x)\cdot (1)\)
 one year ago

CallistoBest ResponseYou've already chosen the best response.0
I'm sorry... I was so blind... Thanks all!!!!
 one year ago

tkhunnyBest ResponseYou've already chosen the best response.1
Don't forget that we DO have to believe that the antiderivative EXISTS! Good work.
 one year ago

CallistoBest ResponseYou've already chosen the best response.0
f(x) = f(x) too, I guess. Since it's an even function..
 one year ago
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