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I need to find the derivative of:

$$ h(x) = \int_{0}^{x^2} (1-t^2)^{1/3} \, dt $$

Would the answer to that just be:

$$ (1-x^4)^{1/3}? $$

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    $h(x) = g(x^2)$ where $g(x) = \int_0^x (1 - t^2)^{1/3} \, dt$. What you've written down is $g'(x^2)$, which is not the same as $h'(x)$. Use the chain rule.2011-05-16
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    See also: http://math.stackexchange.com/questions/6155/derivative-of-integral/6156#61562011-05-16

2 Answers 2