Find the derivatives of $\frac{d}{dx}\int _2^{x^3}\sin \left(t^2\right)dt$ and $\frac{d}{dx}\int _{\tan x}^{\sqrt{x+2}}\sqrt[3]{1+t^2}dt$ I know that $\frac{d}{dx}\int _a^xf\left(t\right)dt=f\left(x\right)\:$ and $\frac{d}{dx}\int _a^{g\left(x\right)}f\left(t\right)dt=f\left(g\left(x\right)\right)\cdot g'\left(x\right)$
I think I know what to do on the first problem, but I'm stumped on the second one. $a=\tan \left(x\right)$ rather than some constant, so I feel like I did not do it correctly.
a. $\frac{d}{dx}\int _2^{x^3}\sin \left(t^2\right)dt$
$$=\sin \left(\left(x^3\right)^2\right)\cdot 3x^2$$
$$=\sin \left(x^6\right)\cdot 3x^2$$
b. $\frac{d}{dx}\int _{\tan x}^{\sqrt{x+2}}\sqrt[3]{1+t^2}dt$
$$=\sqrt[3]{1+\sqrt{x+2}^2}\cdot \frac{1}{2\sqrt{x+2}}$$
$$=\sqrt[3]{x+3}\cdot \frac{1}{2\sqrt{x+2}}$$
$$=\frac{\sqrt[3]{x+3}}{2\sqrt{x+2}}$$
 A: This will be of use:
$$
\frac{d}{dx}\int_{a(x)}^{b(x)}f(t)\;dt = f(b(x))\;b'(x)-f(a(x))\;a'(x)
$$
It's a simplified version of Leibniz Rule https://en.wikipedia.org/wiki/Leibniz_integral_rule. Retry part b using it.
A: This is a standard application of Fundamental Theorem of Calculus.
Since you seen to know what to do for (a), I'll show you the process for (b), although they're near identical.
Let $$F(t) = \int (1+t^2)^{1/3}dt $$
By FTC, 
$$F(\sqrt{x+3}) - F(\tan(x)) = \int_{\tan(x)}^{\sqrt{x+3}} (1+t^2)^{(1/3)}dt$$
Taking the derivatives of both sides, we get
$$\frac{d}{dx} \int_{\tan(x)}^{\sqrt{x+3}} (1+t^2)^{(1/3)} = F'(\sqrt{x+3})\left(\frac{1}{2 \sqrt{x+3}}\right)- F'(\tan(x))(\sec^2(x))$$
And $F'(x) = (1 + x^2)^{1/3}$
I trust you can plug in the rest
A: The Fundamental Theorem of Calculus implies that $$\frac{d}{dx}\int_{g(x)}^{h(x)}f(t)dt = f(h(x))h'(x)-f(g(x))g'(x)$$
So, your first problem is correct only because $$\frac{d}{dx}\int_2^{x^3}\sin \left(t^2\right)dt=\sin((x^3)^2)\cdot3x^2-\sin(2^2)\cdot\underbrace{\frac{d}{dx}(2)}_{=0}$$
and this works when the lower bound is any constant (similar if upper bound is constant).
For the second one, do the same: $$\frac{d}{dx}\int_{\tan(x)}^{\sqrt{x+2}}\sqrt[3]{1+t^2}dt=\sqrt[3]{1+(\sqrt{x+2})^2}\frac{d}{dx}\left(\sqrt{x+2}\right)-\sqrt[3]{1+\tan^2(x)}\cdot\frac{d}{dx}(\tan(x))$$ $$=\frac{\sqrt[3]{x+3}}{2\sqrt{x+2}}-\sqrt[3]{\sec^2(x)}\cdot\sec^2(x)=\frac{\sqrt[3]{x+3}}{2\sqrt{x+2}}-\sec^{8/3}(x)$$
