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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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    "the answer"...to what question?? That can't be the answer ( solution) to the integral you first talk about.2012-08-13
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    what is $y$? Are you saying $y=\int_1^{2^x} t^{1/3} dt$?2012-08-13
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    yes, I meant to say calculate dy/dx.2012-08-13
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    The question as posted *right now* doesn't fit the OP which says literally: "Integral from 1 to 2^x of : (t)^(1/3) How can you calculate dy/dx here" ...why someone decided the lower limit is $\,2^x\,$ is beyond my comprehension but perhaps this illustrates the problem of getting into other people's questions and edit them freely.2012-08-13
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    BTW, and right now again, there are already 5 (five!) edits to this question, which was posted 40 minutes ago...2012-08-13
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    @DonAntonio If you look at the edit history, you will see that the OP himself changed the limits of integration to what they are at the moment. I can only conclude that the phrasing of the original version was wrong, and that the present version reflects the OP's original intention.2012-08-13
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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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    Fundamental theorem of calculus, plus the chain rule...2012-08-13
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    you probably meant $(2^x)^{1/3}$, not just $x^{1/3}$?2012-08-13
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    @gt6989b Of course, thank you!2012-08-13
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    And you can combine terms for $2^\frac{x}{3}2^x$, of course.2012-08-13
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    @ThomasAndrews Of course, thank you! The worst is, even when I teach, I make stupid mistakes like this one...2012-08-13
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    @MTurgeon , don't worry about making stupid mistakes while teaching. Perhaps you can do what I usually do: state at the course's beginning in front of all the students "Guys, you're expected to *think* and deduce stuff on your own in this course. After all, remember that **any** mistake I may do while teaching is your own responsibility". That way they won't be able to say, like little children, "Hey, the teacher said so and so!"2012-08-13
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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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    He had the integral from $1$ to $2^x$, not from $0$.2012-08-13
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    Corrected. Thanks.2012-08-13
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    It's $\int_{2^x}^1$, not $\int_1^{2^x}$.2012-08-13
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    Say who, @S4M?? The ORIGINAL POST says EXACTLY the following: "Integral from 1 to 2^x of : (t)^(1/3) How can you calculate dy/dx here" ...but if people gets into the OP and edits it according to their will then the can actually change deeply the OP!2012-08-13
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    Yeah, the OP changed it.2012-08-13
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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: $\frac{dy}{dx}= -\ln(2)2^{4x/3}$