# Derivatives of Inverse Trigonometric Functions $\frac d {dx} \bigg[\arcsin x\bigg] = \frac 1 {\sqrt {1-x^2}}$ $\frac d {dx} \bigg[\arccos x\bigg] = - \frac 1 {\sqrt {1-x^2}}$ $\frac d {dx} \bigg[ \arctan x\bigg] = \frac 1 {x^2+1}$ ^3wzecb $ \begin{align} \arcsin(\sin\theta) &= \theta \qquad \text {take derivative of both sides} \\ \frac {d } {d\theta} \arcsin(\sin\theta) &= 1 \\ \frac {d(\arcsin (\sin \theta))} {\sin \theta} \cdot \frac {d\sin \theta} {d\theta} &= 1 \\ \frac {d(\arcsin (\sin \theta))} {\sin \theta} \cdot \cos \theta &= 1 \\ \frac {d(\arcsin (\sin \theta))} {\sin \theta} &= \frac 1 {\cos \theta} \\ \frac {d(\arcsin (\sin \theta))} {\sin \theta} &= \frac 1 {\sqrt {1 - \sin^2 \theta}} \\ \text{substitute }x&=\sin \theta \\ \frac {d(\arcsin x)} {dx} &= \frac 1 {\sqrt {1-x^2}} \end{align} $ $ \begin{align} \arccos(\cos\theta) &= \theta \qquad \text {take derivative of both sides} \\ \frac {d } {d\theta} \arccos(\cos\theta) &= 1 \\ \frac {d(\arccos (\cos \theta))} {\cos \theta} \cdot \frac {d\cos \theta} {d\theta} &= 1 \\ \frac {d(\arccos (\cos \theta))} {\cos \theta} \cdot (-\sin \theta) &= 1 \\ \frac {d(\arccos (\cos \theta))} {\cos \theta} &= -\frac 1 {\sin \theta} \\ \frac {d(\arccos (\cos \theta))} {\cos \theta} &= -\frac 1 {\sqrt {1 - \cos^2 \theta}} \\ \text{substitute }x&=\cos \theta \\ \frac {d(\arccos x)} {dx} &= -\frac 1 {\sqrt {1-x^2}} \end{align} $ $ \begin{align} \arctan(\tan\theta) &= \theta \qquad \text {take derivative of both sides} \\ \frac {d } {d\theta} \arctan(\tan\theta) &= 1 \\ \frac {d(\arctan (\tan \theta))} {\tan \theta} \cdot \frac {d\tan \theta} {d\theta} &= 1 \\ \frac {d(\arctan (\tan \theta))} {\tan \theta} \cdot \sec^2 \theta &= 1 \\ \frac {d(\arctan (\tan \theta))} {\tan \theta} &= \frac 1 {\sec^2 \theta} \\ \frac {d(\arctan (\tan \theta))} {\tan \theta} &= \frac 1 {1 + \tan^2 \theta} \\ \text{substitute }x&=\tan \theta \\ \frac {d(\arctan x)} {dx} &= \frac 1 {1+x^2} \end{align} $