Question

Can anyone explain me how to solve this? \[\frac{ d }{ dx }\int\limits_{2\Pi}^{\sqrt[3]{x}} \sin(t)dt=\]

Answer 1

Apply the fundamental theorem of calculus:
\[\large \frac{d}{dx}\int_c^{g(x)}f(t)~dt=f(g(x))\cdot g'(x)\]
To see why this works (at least for this example), you can simply compute the integral, then differentiate:
\[\Large\begin{align*}\int_{2\pi}^\sqrt[3]x\sin t~dt&=-\left[\cos t\right]_{2\pi}^{\sqrt[3]x}\\
&=-\cos\sqrt[3]x-(-\cos2\pi)\\
&=-\cos\sqrt[3]x+1
\end{align*}\]
Differentiating gives
\[\large\begin{align*}\frac{d}{dx}[\cdots]&=-\sin\sqrt[3]x\cdot\frac{d}{dx}[\sqrt[3]x]\\
&=\frac{1}{3}\sin\sqrt[3]x \cdot\frac{1}{x^{2/3}}
\end{align*}\]

Answer 2

Notice that in this case, \(g(x)=\sqrt[3]x\) and \(f(t)=\sin t\), which indeed gives
\[\large f(g(x))=\sin\sqrt[3]x\]
and \(g'(x)=\dfrac{1}{3x^{2/3}}\).

Answer 3

wait what happens to the +1? before differentiating.

Answer 4

@SithsAndGiggles

Answer 5

What's the derivative of a constant?

Answer 6

got it. I was anti deriving