Question: Suppose $F(x) = \int^{x^2}_0 \frac{1}{\cos t} dt$. Find the derivative of $F(x)$ over the region $x\in[0, \frac{\pi}{4}]$ for which it is continuous.
Attempt: So I know how to do the following computations:
$$\int^{x^2}_0 \frac{1}{\cos(t)} dt = \int^{x^2}_0 \sec(t)dt$$
$$\int^{x^2}_0 \frac{1}{\cos(t)} dt = [\ln|\tan(t)+\sec(t)|]^{x^2}_0$$
$$\int^{x^2}_0 \frac{1}{\cos(t)} dt = \ln|\tan(x^2)+\sec(x^2)| – \ln|\tan(0)+\sec(0)|$$
$$\int^{x^2}_0 \frac{1}{\cos(t)} dt = \ln|\tan(x^2)+\sec(x^2)| – \ln1$$
$$\int^{x^2}_0 \frac{1}{\cos(t)} dt = \ln|\tan(x^2)+\sec(x^2)|$$
Note: To find $\int \sec(t) dt$ I used the method of substitution found here.
And then using the chain rule to differentiate ($u=|\tan(x^2)+\sec(x^2)|$ so $F(x)=\ln u$), we get $F'(x)=\frac{2x}{\cos (x^2)} = 2x\sec (x^2)$ for $x\in[0, \frac{\pi}{4}]$.
However, this question was given in a real analysis course and I believe I need to use the Fundamental Theorem of Calculus (which is fine considering $F$ is continuous for $x\in[0, \frac{\pi}{4}]$ as stated in the question so not required to prove).
I am familiar with two different (but equivalent) definitions of the FTC; this one and this one, but am unsure of which one to use.
Any help would be greatly appreciated.
Best Answer
$\newcommand{\d}[1]{\; \mathrm{d} #1}$ Write: $$ G(x) = \int_0^x \sec{t} \d{t} $$ FTOC then tells us that $G'(x) = \sec{x}$. Furthermore, by chain rule we know that: $$ (f \circ g)' (x) = g'(x) f'(g(x)) $$ Therefore, let $H(x) = x^2$, and we have that: \begin{align*} \frac{\mathrm{d}}{\mathrm{d}x} \int_0^{x^2} \sec{t} \d{t} &= (G \circ H)'(x) \\ &= H'(x)G'(H(x)) \\ &= 2x\sec{x^2} \end{align*}