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imagreencatBest ResponseYou've already chosen the best response.2
You just need to replace all zs with x. Since you're getting the derivative of the integral. Also, you don't need to multiply anything more, because the derivative of x is just 1. But, let's say the upper bound wasn't just x but a function that is differentiable, say sin x. There is a need to multiply an additional cos x after replacing all zs with sin x.
 one year ago

rulnickBest ResponseYou've already chosen the best response.0
I agree with imagreencat. The answer is F'(x) = ( cos x ) / (x^3 + 1).
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
Let, \( \LARGE g(z) = \frac{(\cos z)}{(z^3 +1)} \) and \( G(z) = \int g(z) dz\) Then, \( \LARGE F(x) = \int_{0}^{x} \frac{(cosz)}{(z^3 +1)} dz = G(x)  G(0)\) \( \LARGE F'(x) = G'(x)  0 = \frac{ ( cos x ) }{(x^3 + 1).}\)
 one year ago
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