### Derivatives of Inverse Trigonometric Functions Inverse trigonometric functions (e.g., $\arcsin x$, $\arccos x$, $\arctan x$) frequently appear in calculus problems involving angles, rates of change, or integrals. Their derivatives are essential for solving optimization, related rates, and differential equations. #### Key Derivatives and Proofs **1. $\frac{d}{dx} \arcsin x$**: Set $y = \arcsin x$, so $x = \sin y$. Differentiate implicitly: $$ \frac{d}{dx}(x) = \frac{d}{dx}(\sin y) \implies 1 = \cos y \cdot \frac{dy}{dx} $$ Solve for $\frac{dy}{dx}$: $$ \frac{dy}{dx} = \frac{1}{\cos y} $$ Using $\cos y = \sqrt{1 - \sin^2 y} = \sqrt{1 - x^2}$ (since $y \in [-\frac{\pi}{2}, \frac{\pi}{2}]$, $\cos y \geq 0$): $$ \boxed{\frac{d}{dx} \arcsin x = \frac{1}{\sqrt{1 - x^2}} \quad (-1 < x < 1) $$ **2. $\frac{d}{dx} \arccos x$**: Similarly, let $y = \arccos x$, so $x = \cos y$. Implicit differentiation gives: $$ 1 = -\sin y \cdot \frac{dy}{dx} \implies \frac{dy}{dx} = -\frac{1}{\sin y} = -\frac{1}{\sqrt{1 - x^2}} $$ $$ \boxed{\frac{d}{dx} \arccos x = -\frac{1}{\sqrt{1 - x^2}} \quad (-1 < x < 1) $$ **3. $\frac{d}{dx} \arctan x$**: Let $y = \arctan x$, so $x = \tan y$. Differentiate implicitly: $$ 1 = \sec^2 y \cdot \frac{dy}{dx} \implies \frac{dy}{dx} = \cos^2 y $$ Since $\cos y = \frac{1}{\sqrt{1 + \tan^2 y}} = \frac{1}{\sqrt{1 + x^2}}$: $$ \boxed{\frac{d}{dx} \arctan x = \frac{1}{1 + x^2} \quad (-\infty < x < \infty) $$ #### Derivatives of Other Inverse Trig Functions - $\frac{d}{dx} \arcsec x = \frac{1}{|x|\sqrt{x^2 - 1}} \quad (|x| > 1)$ - $\frac{d}{dx} \arccsc x = -\frac{1}{|x|\sqrt{x^2 - 1}} \quad (|x| > 1)$ - $\frac{d}{dx} \arccot x = -\frac{1}{1 + x^2} \quad (-\infty < x < \infty)$ **Notes**: - The absolute values ($|x|$) in $\arcsec$ and $\arccsc$ ensure correct sign handling across domains. - $\arccot x$’s derivative is negative of $\arctan x$’s, reflecting their co-function relationship. #### Examples **Example 1**: Find $\frac{d}{dx} \arctan(3x^2)$. Apply the chain rule: $$ \frac{d}{dx} \arctan(3x^2) = \frac{1}{1 + (3x^2)^2} \cdot 6x = \frac{6x}{1 + 9x^4} $$ **Example 2**: Differentiate $y = \arcsin(\sqrt{x})$. $$ \frac{dy}{dx} = \frac{1}{\sqrt{1 - (\sqrt{x})^2}} \cdot \frac{d}{dx}(\sqrt{x}) = \frac{1}{\sqrt{1 - x}} \cdot \frac{1}{2\sqrt{x}} = \frac{1}{2\sqrt{x - x^2}} \quad (0 < x < 1) $$ **Table Summary**: | **Function** | **Derivative** | |--------------|----------------| | $\arcsin x$ | $\frac{1}{\sqrt{1 - x^2}}$ | | $\arccos x$ | $-\frac{1}{\sqrt{1 - x^2}}$ | | $\arctan x$ | $\frac{1}{1 + x^2}$ | | $\arcsec x$ | $\frac{1}{|x|\sqrt{x^2 - 1}}$ | | $\arccsc x$ | $-\frac{1}{|x|\sqrt{x^2 - 1}}$ | | $\arccot x$ | $-\frac{1}{1 + x^2}$ |

1: Functions and graphs

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