Recall $$\arcsin (x) = \theta$$ means that $\sin (\theta ) = x$ and $\answer [given]{\frac {-\pi }{2}}\le \theta \le \answer [given]{\frac {\pi }{2}}$. Implicitly differentiating with respect $x$ we see $$\begin{align*} \sin(\theta) &= x\\ \frac{d}{dx} \sin(\theta) &= \frac{d}{dx} x &\text{Differentiate both sides.}\\ \cos(\theta) \cdot \theta' &= 1 &\text{Implicit differentiation.}\\ \theta' &= \frac{1}{\cos(\theta)} &\text{Solve for $\theta'$.} \end{align*}$$ While $\theta ' = \frac {1}{\cos (\theta )}$, we need our answer written in terms of $x$. Since we are assuming that $$\sin (\theta ) = x,$$ consider the following triangle with the unit circle: From the unit circle above, we see that $$\begin{align*} \theta' &= \frac{1}{\cos(\theta)}\\ &=\frac{\mathrm{hyp}}{\mathrm{adj}}\\ &= \answer[given]{\frac{1}{\sqrt{1-x^2}}}. \end{align*}$$ Recall, $\cos {x}\ge 0$ on the interval $\Bigl (-\frac {\pi }{2},\frac {\pi }{2}\Bigr )$.

To be completely explicit, $$\frac {d}{dx} \theta = \frac {d}{dx} \arcsin (x) = \answer [given]{\frac {1}{\sqrt {1-x^2}}}.$$

To start, note that the Inverse Function Theorem assures us that this derivative actually exists. Recall $$\arctan (x) = \theta$$ means that $\tan (\theta ) = x$ and $\answer [given]{\frac {-\pi }{2}}< \theta < \answer [given]{\frac {\pi }{2}}$.

Implicitly differentiating with respect to $x$ we obtain that

$$\begin{align*} \tan(\theta) &= x\\ \frac{d}{dx} \tan(\theta) &= \frac{d}{dx} x &\text{Differentiate both sides.}\\ \sec^2(\theta) \cdot \theta' &= 1 &\text{Implicit differentiation.}\\ \theta' &= \frac{1}{\sec^2(\theta)} \end{align*}$$ Recall the trig identity: $\sec ^2(\theta )=1+\tan ^2(\theta )$.
$$\begin{align*} \theta' &= \frac{1}{1+\tan^2(\theta)} \end{align*}$$ Recall from above: $\tan (\theta ) = x$.
$$\begin{align*} \theta' &= \frac{1}{1+x^2} \end{align*}$$

To be completely explicit, $$\frac {d}{dx} \theta = \frac {d}{dx} \arctan (x) = \answer [given]{\frac {1}{1+x^2}}.$$

Recall $$\text {arcsec(x)} = \theta$$ means that $\sec (\theta ) = x$ and $\answer [given]{0}\le \theta \le \answer [given]{\pi }$ with $\theta \ne \answer [given]{\pi /2}$. Implicitly differentiating with respect $x$ we see $$\begin{align*} \sec(\theta) &= x\\ \frac{d}{dx} \sec(\theta) &= \frac{d}{dx} x &\text{Differentiate both sides.}\\ \sec(\theta)\tan(\theta) \cdot \theta' &= 1 &\text{Implicit differentiation.}\\ \theta' &= \frac{1}{\sec(\theta)\tan(\theta)} &\text{Solve for $\theta'$.}\\ \theta' &= \frac{\cos^2(\theta)}{\sin(\theta)}. \end{align*}$$ While $\theta ' = \frac {\cos ^2(\theta )}{\sin (\theta )}$, we need our answer written in terms of $x$. Since we are assuming that $$\sec (\theta ) = \frac {1}{\cos (\theta )}= x,$$ we may consider the following triangle: We may now scale this triangle by a factor of $\frac {1}{x}$ to place it on the unit circle: From the unit circle above, we see that $$\begin{align*} \theta' &= \frac{\cos^2(\theta)}{\sin(\theta)}\\ &= \frac{ \left(\frac{\mathrm{adj}}{\mathrm{hyp}}\right)^2}{\frac{\mathrm{opp}}{\mathrm{hyp}}}\\ &= \frac{\left(\mathrm{adj}\right)^2}{\mathrm{opp}} &\text{Note, $\mathrm{hyp}=1$.}\\ &= \frac{\answer[given]{1/x^2}}{\sqrt{1-1/x^2}}\\ &= \frac{1}{|x|\sqrt{x^2-1}}, \end{align*}$$ To be completely explicit, $$\frac {d}{dx} \theta = \frac {d}{dx} \text {arcsec(x)} = \frac {1}{|x|\sqrt {x^2-1}}\qquad \text {for $|x|>1$}.$$