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  • 4.9: Inverse Trigonometric Functions
    https://math.libretexts.org/Bookshelves/Calculus/Calculus_(Guichard)/04%3A_Transcendental_Functions/4.09%3A_Inverse_Trigonometric_Functions
    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 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 .
  • 2.7: Derivatives of Inverse Functions
    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...\[\begin{align} &1. \frac{d}{dx}\big(\sin^{-1}(x)\big) = \frac{1}{\sqrt{1-x^2}} \qquad &&4.\frac{d}{dx}\big(\cos^{-1}(x)\big) = -\frac{1}{\sqrt{1-x^2}} \\ &2.\frac{d}{dx}\big(\sec^{-1}(x)\big) = \frac{1}{|x|\sqrt{x^2-1}} &&5.\frac{d}{dx}\big(\csc^{-1}(x)\big) = -\frac{1}{|x|\sqrt{x^2-1}} \\ &3.\frac{d}{dx}\big(\tan^{-1}(x)\big) = \frac{1}{1+x^2} &&6.\frac{d}{dx}\big(\cot^{-1}(x)\big) = -\frac{1}{1+x^2} \end{align}\]

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