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squirtskid

  • 4 years ago

Find F'(x) if F(x) = integral from 0 to x of ((cosz)/(z^3 +1)

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  1. imagreencat
    • 4 years ago
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    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.

  2. rulnick
    • 4 years ago
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    I agree with imagreencat. The answer is F'(x) = ( cos x ) / (x^3 + 1).

  3. experimentX
    • 4 years ago
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    Let, \( \LARGE g(z) = \frac{(\cos z)}{(z^3 +1)} \) and \( G(z) = \int g(z) dz\) Then, \( \LARGE F(x) = \int_{0}^{x} \frac{(cosz)}{(z^3 +1)} dz = G(x) - G(0)\) \( \LARGE F'(x) = G'(x) - 0 = \frac{ ( cos x ) }{(x^3 + 1).}\)

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