Inverse Trigonometric Functions
For background on the inverse functions and how the inverse trig functions are defined, you can refer to my series on trigonometry and this post.
All proofs will be using the inverse derivative property. These results are very commonly used in advanced integration.
Inverse Sine
\[\begin{align}
\frac{d}{dx} \arcsin (x)
&= \frac{1}{\frac{d}{dz} \sin(z) \bigg\rvert_{z=\arcsin (x)} } \\[10pt]
&= \frac{1}{\cos(z) \bigg\rvert_{z=\arcsin (x)} } \\[10pt]
&= \frac{1}{\cos(\arcsin (x))} \\[10pt]
&= \frac{1}{\sqrt{\cos^2(\arcsin (x))}} \\[10pt]
&= \frac{1}{\sqrt{1 - \sin^2(\arcsin (x))}} \\[10pt]
&= \frac{1}{\sqrt{1 - x^2}}
\end{align}\]
Inverse Cosine
\[\begin{align}
\frac{d}{dx} \arccos (x)
&= \frac{1}{\frac{d}{dz} \cos(z) \bigg\rvert_{z=\arccos (x)} } \\[10pt]
&= \frac{1}{-\sin(z) \bigg\rvert_{z=\arccos (x)} } \\[10pt]
&= \frac{-1}{\sin(\arccos (x))} \\[10pt]
&= \frac{-1}{\sqrt{\sin^2(\arccos (x))}} \\[10pt]
&= \frac{-1}{\sqrt{1 - \cos^2(\arccos (x))}} \\[10pt]
&= \frac{-1}{\sqrt{1 - x^2}}
\end{align}\]
Inverse Tangent
\[\begin{align}
\frac{d}{dx} \arctan (x)
&= \frac{1}{\frac{d}{dz} \tan(z) \bigg\rvert_{z=\arctan (x)} } \\[10pt]
&= \frac{1}{\sec^2(z) \bigg\rvert_{z=\arctan (x)} } \\[10pt]
&= \frac{1}{\sec^2(\arctan (x))} \\[10pt]
&= \frac{1}{1 + \tan^2(\arctan (x))} \\[10pt]
&= \frac{1}{1 + x^2}
\end{align}\]
Inverse Secant
\[\begin{align}
\frac{d}{dx} \arcsec (x)
&= \frac{1}{\frac{d}{dz} \sec(z) \bigg\rvert_{z=\arcsec (x)} } \\[10pt]
&= \frac{1}{\sec(z) \tan(z) \bigg\rvert_{z=\arcsec (x)} } \\[10pt]
&= \frac{1}{\sec(\arcsec (x)) \tan(\arcsec (x)) } \\[10pt]
&= \frac{1}{x \sqrt{\tan^2(\arcsec (x))} } \\[10pt]
&= \frac{1}{x \sqrt{1 + \sec^2(\arcsec (x))} } \\[10pt]
&= \frac{1}{x \sqrt{1 + x^2} } \\[10pt]
\end{align}\]
Inverse Cosecant
\[\begin{align}
\frac{d}{dx} \arccsc (x)
&= \frac{1}{\frac{d}{dz} \csc(z) \bigg\rvert_{z=\arccsc (x)} } \\[10pt]
&= \frac{1}{- \csc(z) \cot(z) \bigg\rvert_{z=\arccsc (x)} } \\[10pt]
&= \frac{-1}{\csc(\arccsc (x)) \cot(\arccsc (x)) } \\[10pt]
&= \frac{-1}{x \sqrt{\cot^2(\arccsc (x))} } \\[10pt]
&= \frac{-1}{x \sqrt{1 + \csc^2(\arccsc (x))} } \\[10pt]
&= \frac{- 1}{x \sqrt{1 + x^2} } \\[10pt]
\end{align}\]
Inverse Cotangent
\[\begin{align}
\frac{d}{dx} \arccot (x)
&= \frac{1}{\frac{d}{dz} \cot(z) \bigg\rvert_{z=\arccot (x)} } \\[10pt]
&= \frac{1}{- \csc^2(z) \bigg\rvert_{z=\cot^{-1} (x)} } \\[10pt]
&= \frac{-1}{\csc^2(\cot^{-1} (x))} \\[10pt]
&= \frac{-1}{1 + \cot^2(\cot^{-1} (x))} \\[10pt]
&= \frac{-1}{1 + x^2}
\end{align}\]