Find the derivative of the following with $x \in [\frac{\pi}{4},\frac{\pi}{2}]$:
$$H(x)= \int_0^x \frac{1}{\sin(t)}dt$$ and $$I(x)= \int_0^{x^2}2^sds$$
I said that since both of the functions are continuous on $[\frac{\pi}{4},\frac{\pi}{2}]$, both of the functions inside the integral are Riemann integrable, by the Fundamental theorem of Calculus, and hence $H(x)$ and $I(x)$ are continuous on the same interval and $H'(x) = \frac{1}{\sin(x)} $ and $I'(x) = 2^{x^2}$
Is this correct?
Note that $$\int_0^{\pi/4} \frac{1}{\sin t}dt \ge \int_0^{\pi/4} \frac{1}{x}dx \to \text{ diverges}$$
Nor is correct the second.$$I(x) = \int_0^{x^2} 2^sds = \mathcal I(x^2)$$ where $\mathcal I(x) := \int_0^x 2^s ds$. Then, since $\mathcal I'(x) = 2^x$, we have $$I'(x) = \left[\mathcal I(x^2)\right]' = 2x\cdot\mathcal I'(x^2) = 2x\cdot 2^{x^2}$$
Also, see Leibniz's rule for more information.