# Rules of Differentiation of Inverse Trigonometric Functions

 
 In what follows, $u$ is a function of $x$. Basic Inverse Trigonometric Functions $\dfrac{d }{d x}\sin^{-1} x = \dfrac{1}{\sqrt{1-x^2}}$ $\dfrac{d }{d x}\cos^{-1} x= \dfrac{-1}{\sqrt{1-x^2}}$ $\dfrac{d }{d x}\tan^{-1} x= \dfrac{1}{1+x^2}$ $\dfrac{d }{d x}\cot^{-1} x= \dfrac{-1}{1+x^2}$ Composite Inverse Trigonometric Functions $\dfrac{d }{d x}\sin^{-1} u= \dfrac{1}{\sqrt{1-u^2}} \dfrac{d u}{d x}$ Example $\dfrac{d }{d x}\sin^{-1} (x^3-6) = \dfrac{1}{\sqrt{1-(x^3-6)^2}} (3x^2)$ $=\dfrac{3x^2}{\sqrt{1-(x^3-6)^2}}$ $\dfrac{d }{d x}\cos^{-1} u= \dfrac{-1}{\sqrt{1-u^2}} \dfrac{d u}{d x}$ Example $\dfrac{d }{d x}\cos^{-1} (-3x+7)= \dfrac{-1}{\sqrt{1-(-3x+7)^2}} (-3)$ $= \dfrac{3}{\sqrt{1-(-3x+7)^2}}$ $\dfrac{d }{d x}\tan^{-1} u= \dfrac{1}{1+u^2} \dfrac{d u}{d x}$ Example $\dfrac{d }{d x}\tan^{-1} (4x^4-2) = \dfrac{1}{1+(4x^4-2)^2} (16x^3)$ $= \dfrac{16x^3}{1+(4x^4-2)^2}$ $\dfrac{d }{d x}\cot^{-1} u= \dfrac{-1}{1+u^2} \dfrac{d u}{d x}$ Example $\dfrac{d }{d x}\cot^{-1} (-4x+3) = \dfrac{-1}{1+(-4x+3)^2} (-4)$ $=\dfrac{4}{1+(-4x+3)^2}$