Recall 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

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,

To start, note that the Inverse Function Theorem assures us that this derivative actually exists. Recall 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,

Recall 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

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,