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How can you calculate $dy/dx$ here?

$y=\int_{2^x}^{1}t^{1/3}dt$

I get that the anti derivative is $3/4t^{4/3}$, but I don't understand what I'm supposed to do next.

The answer is

$\int_x^1\sqrt{1-t^2}dt + 2$

I have no idea how to get there

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    @MTurgeon, that seems to be accurate. Still, the issue of ever-changing questions, edited either by the OP or by someone else, is imo annoying, but alas it's something we can't do anything about, apparently. Thanks.2012-08-13

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What about using the Fundamental Theorem of Calculus?

$\begin{align*}\dfrac{dy}{dx} &= \dfrac{d}{dx}\left(\int_{2^x}^1t^{1/3}dt\right)\\ &= - \dfrac{d}{dx}\left(\int^{2^x}_1t^{1/3}dt\right)\\ &= -((2^x)^{1/3})\dfrac{d}{dx}(2^x)\\ &= -2^{4x/3}\ln 2. \end{align*}$

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    @DonAntonio Thanks for the tip!2012-08-13
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$y:=\int_1^{2^x}t^{1/3}dt=\left.\frac{3}{4}t^{4/3}\right|_1^{2^x}=\frac{3}{4}\left[(2^x)^{4/3}-1\right]=\frac{3}{4}(2^{4x/3}-1)$

Added: $\;\;\frac{dy}{dx}=\frac{3}{4}\frac{4}{3}2^{4x/3}\log 2=2^{4x/3}\log 2$

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    Seems so, @ThomasAndrews, and this is annoying as sometimes the edited question can be pretty different from the new one. I'll leave my answer as it is just the same but with the integral's limits interchanged.2012-08-13
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If you call $\varphi(z)=\int_1^z t^{1/3}dt$, then you have $y=-\varphi(2^x)$, then $\frac{dy}{dx}=-\ln(2)2^x \varphi'(2^x) = -\ln(2)2^x (2^x)^{1/3}$

So: $\frac{dy}{dx}= -\ln(2)2^{4x/3}$