derivatives of inverse trigonometric functions

Let \[y=\sin^{-1}x\], where \[-\frac{\pi}{2}\le y\le \frac{\pi}{2}\]. Thus, \[x=\sin y\].
Now to find \[y^{\prime}\], we will use implicit differentiation. Since \[\sin y=x\], then \[(\sin y)^{\prime}=x^{\prime}\]. Another equation we have to establish is that since \[\sin^{2}\theta+\cos^{2}\theta=1\], thus \[\cos\theta=\sqrt{1-\sin^{2}\theta}\].

\begin{align*} (\sin y)^{\prime}&=x^{\prime}\\ \cos y\cdot y^{\prime}&=1\\ \sqrt{1-\sin^{2}y}\cdot y^{\prime}&=1\\ \\ \text{Substituting }\sin y&=x,\\ y^{\prime}&=\frac{1}{\sqrt{1-x^{2}}}\\ \end{align*}

Similar to proving \[\left( \sin^{-1}x \right)^{\prime}\], we let \[y=\cos^{-1}x\], thus \[\cos y=x\], where \[0\le y\le\pi\].

\begin{align*} (\cos y)^{\prime}&=x^{\prime}\\ -\sin y\cdot y^{\prime}&=1\\ -\sqrt{1-\cos^{2}y}\cdot y^{\prime}&=1\\ y^{\prime}&=-\frac{1}{\sqrt{1-x^{2}}}\\ \end{align*}

Let \[y=\tan^{-1}x\], thus \[\tan y=x\], where \[-\frac{\pi}{2}\le y\le \frac{\pi}{2}\].

\begin{align*} (\tan y)^{\prime}&=x^{\prime}\\ \sec^{2}y\cdot y^{\prime}&=1\\ \left( 1+\tan^{2}y \right)\cdot y^{\prime}&=1\\ y^{\prime}&=\frac{1}{1+x^{2}}\\ \end{align*}

Let \[y=\csc^{-1}x\], thus \[\csc y=x\], where \[y\in\left\{y\mid-\frac{\pi}{2}\le y\le \frac{\pi}{2},y\ne0\right\}\] and \[x\in\{x\mid |x|\ge 1\}\].

\begin{align*} (\csc y)^{\prime}&=x^{\prime}\\ -\csc y\cot y\cdot y^{\prime}&=1\\ -x \sqrt{\csc^{2}-1}\cdot y^{\prime}&=1\\ y^{\prime}&=-\frac{1}{x \sqrt{x^{2}-1}}\\ &=-\frac{1}{|x|\sqrt{x^{2}-1}}\\ \end{align*}

Since \[\csc y\cot y\] is always positive in the interval of \[y\] (see trigonometry graphs for visual proof), as when \[\csc y\] is negative, \[\cot y\] is also negative, thus making it positive, and \[\sqrt{x^{2}-1}\] is also always positive, thus \[x\] must always be positive. But since the range of \[x\] based on the graph is \[x=\left(-\infty,-1\right]\cup \left[1,+\infty\right)\], we must limit \[x\] to only positive, thus \[y^{\prime}=-\frac{1}{|x|\sqrt{x^{2}-1}}\].
Another explanation would be the gradient of the graph \[\csc^{-1}x\] is always negative, thus \[x \sqrt{x^{2}-1}\] must always be positive, so that \[y^{\prime}\] stays negative. A third way to prove \[|x|\] would be by proving the derivative of \[\csc^{-1}x\] with chain rule.

Let \[y=\sec^{-1}x\], thus \[\sec y=x\], where \[y\in\left\{ y\mid 0\le y<\frac{\pi}{2},\frac{\pi}{2}< y\le \pi \right\}\] and \[x\in \{x\mid |x|\ge1\}\].

\begin{align*} (\sec y)^{\prime}&=x^{\prime}\\ (\sec y\tan y)\cdot y^{\prime}&=1\\ (\sec y\sqrt{\sec^{2}y-1})\cdot y^{\prime}&=1\\ y^{\prime}&=\frac{1}{x \sqrt{x^{2}-1}}\\ &=\frac{1}{|x|\sqrt{x^{2}-1}}\\ \end{align*}

Since \[\sec y\tan y\] is always positive in the interval of \[y\], thus similar to the explanation for \[\csc^{-1}x\], \[x\] must always be positive.

Let \[y=\cot^{-1}x\], thus \[\cot y=x\], where \[y\in \{y\mid 0< y< \pi\}\].

\begin{align*} (\cot y)^{\prime}&=x^{\prime}\\ -\csc^{2}y\cdot y^{\prime}&=1\\ -(1+\cot^{2}y)\cdot y^{\prime}&=1\\ y^{\prime}&=-\frac{1}{(1+x^{2})}\\ \end{align*}