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- https://math.libretexts.org/Bookshelves/Calculus/Calculus_(Guichard)/04%3A_Transcendental_Functions/4.09%3A_Inverse_Trigonometric_FunctionsThe trigonometric functions frequently arise in problems, and often it is necessary to invert the functions, for example, to find an angle with a specified sine. Of course, there are many angles with ...The trigonometric functions frequently arise in problems, and often it is necessary to invert the functions, for example, to find an angle with a specified sine. Of course, there are many angles with the same sine, so the sine function doesn't actually have an inverse that reliably "undoes'' the sine function. If you know that sin x=0.5, you can't reverse this to discover x as there are infinitely many angles with sine 0.50.5 .
- https://math.libretexts.org/Bookshelves/Calculus/Calculus_by_David_Guichard_(Improved)/04%3A_Transcendental_Functions/4.09%3A_Inverse_Trigonometric_FunctionsThe trigonometric functions frequently arise in problems, and often it is necessary to invert the functions, for example, to find an angle with a specified sine. Of course, there are many angles with ...The trigonometric functions frequently arise in problems, and often it is necessary to invert the functions, for example, to find an angle with a specified sine. Of course, there are many angles with the same sine, so the sine function doesn't actually have an inverse that reliably "undoes'' the sine function. If you know that sin x=0.5, you can't reverse this to discover x as there are infinitely many angles with sine 0.50.5 .
- https://math.libretexts.org/Bookshelves/Calculus/Calculus_3e_(Apex)/02%3A_Derivatives/2.07%3A_Derivatives_of_Inverse_Functions\[\begin{align} &1. \frac{d}{dx}\big(\sin^{-1}(x)\big) = \frac{1}{\sqrt{1-x^2}} \qquad &&4.\frac{d}{dx}\big(\cos^{-1}(x)\big) = -\frac{1}{\sqrt{1-x^2}} \\ &2.\frac{d}{dx}\big(\sec^{-1}(x)\big) = \frac...\[\begin{align} &1. \frac{d}{dx}\big(\sin^{-1}(x)\big) = \frac{1}{\sqrt{1-x^2}} \qquad &&4.\frac{d}{dx}\big(\cos^{-1}(x)\big) = -\frac{1}{\sqrt{1-x^2}} \\ &2.\frac{d}{dx}\big(\sec^{-1}(x)\big) = \frac{1}{|x|\sqrt{x^2-1}} &&5.\frac{d}{dx}\big(\csc^{-1}(x)\big) = -\frac{1}{|x|\sqrt{x^2-1}} \\ &3.\frac{d}{dx}\big(\tan^{-1}(x)\big) = \frac{1}{1+x^2} &&6.\frac{d}{dx}\big(\cot^{-1}(x)\big) = -\frac{1}{1+x^2} \end{align}\]
