Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
    
"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
1  
@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
show 2 more comments

3 Answers 3

up vote 4 down vote accepted

What about using the Fundamental Theorem of Calculus?

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

share|improve this answer
1  
Fundamental theorem of calculus, plus the chain rule... –  Thomas Andrews Aug 13 '12 at 14:42
2  
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
1  
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
show 2 more comments

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

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

share|improve this answer
    
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
show 1 more comment

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

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.