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. $\frac{d}{dx}\int _{\tan x}^{\sqrt{x+2}}\sqrt[3]{1+t^2}dt$
$$=\sqrt[3]{1+\sqrt{x+2}^2}\cdot \frac{1}{2\sqrt{x+2}}$$
$$=\sqrt[3]{x+3}\cdot \frac{1}{2\sqrt{x+2}}$$
$$=\frac{\sqrt[3]{x+3}}{2\sqrt{x+2}}$$

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: $$\frac{d}{dx}\int_{\tan(x)}^{\sqrt{x+2}}\sqrt[3]{1+t^2}dt=\sqrt[3]{1+(\sqrt{x+2})^2}\frac{d}{dx}\left(\sqrt{x+2}\right)-\sqrt[3]{1+\tan^2(x)}\cdot\frac{d}{dx}(\tan(x))$$ $$=\frac{\sqrt[3]{x+3}}{2\sqrt{x+2}}-\sqrt[3]{\sec^2(x)}\cdot\sec^2(x)=\frac{\sqrt[3]{x+3}}{2\sqrt{x+2}}-\sec^{8/3}(x)$$
