# How do you calculate the derivative of an integral

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.

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

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. – DonAntonio Aug 13 '12 at 14:24
what is $y$? Are you saying $y=\int_1^{2^x} t^{1/3} dt$? – gt6989b Aug 13 '12 at 14:26
yes, I meant to say calculate dy/dx. – mr real lyfe Aug 13 '12 at 14:28
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. – DonAntonio Aug 13 '12 at 14:57
@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. – DonAntonio Aug 13 '12 at 18:01

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... – Thomas Andrews Aug 13 '12 at 14:42
you probably meant $(2^x)^{1/3}$, not just $x^{1/3}$? – gt6989b Aug 13 '12 at 14:42
@gt6989b Of course, thank you! – M Turgeon Aug 13 '12 at 14:47
And you can combine terms for $2^\frac{x}{3}2^x$, of course. – Thomas Andrews Aug 13 '12 at 15:07
@ThomasAndrews Of course, thank you! The worst is, even when I teach, I make stupid mistakes like this one... – M Turgeon Aug 13 '12 at 15:10

$$y:=\int_1^{2^x}t^{1/3}dt=\left.\frac{3}{4}t^{4/3}\right|_1^{2^x}=\frac{3}{4}\left[(2^x)^{4/3}-1\right]=\frac{3}{4}(2^{4x/3}-1)$$

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 $0$. – Thomas Andrews Aug 13 '12 at 14:30
Corrected. Thanks. – DonAntonio Aug 13 '12 at 14:46
It's $\int_{2^x}^1$, not $\int_1^{2^x}$. – S4M Aug 13 '12 at 14:51
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! – DonAntonio Aug 13 '12 at 14:53
Yeah, the OP changed it. – Thomas Andrews Aug 13 '12 at 15:06

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

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