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
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
What about using the Fundamental Theorem of Calculus?
$\begin{align*}\dfrac{dy}{dx} &= \dfrac{d}{dx}\left(\int_{2^x}^1t^{1/3}dt\right)\\ &= - \dfrac{d}{dx}\left(\int^{2^x}_1t^{1/3}dt\right)\\ &= -((2^x)^{1/3})\dfrac{d}{dx}(2^x)\\ &= -2^{4x/3}\ln 2. \end{align*}$
$y:=\int_1^{2^x}t^{1/3}dt=\left.\frac{3}{4}t^{4/3}\right|_1^{2^x}=\frac{3}{4}\left[(2^x)^{4/3}-1\right]=\frac{3}{4}(2^{4x/3}-1)$
Added: $\;\;\frac{dy}{dx}=\frac{3}{4}\frac{4}{3}2^{4x/3}\log 2=2^{4x/3}\log 2$
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: $\frac{dy}{dx}= -\ln(2)2^{4x/3}$