📂Calculus

Differentiation of Inverse Trigonometric Functions

Theorem¹

$$\begin{align*} \left( \sin^{-1}x \right)^{\prime} &= {{1} \over {\sqrt{1-x^2}}} \\ \left( \cos^{-1}x \right)^{\prime} &= -{{1} \over {\sqrt{1-x^2}}} \\ \left( \tan^{-1}x \right)^{\prime} &= {{1} \over {1+x^2}} \end{align*}$$

Explanation

They are read as arcsine, arccosine, and arctangent, respectively. It might seem surprising that these can be differentiated, but it turns out to be quite simple. As one can see on the right side, the shapes of the derivatives are not that unfamiliar or complicated. Even though they seem utterly unrelated to trigonometric functions, they are widely used, so it’s advisable to memorize their proofs. Details like domains should be carefully considered on one’s own. The proof method itself doesn’t significantly differ whether it concerns cosine or tangent, so we will only cover arcsine $\sin ^{-1}$ here.

Proof

Let’s assume $y=\sin^{-1} x$. Since $\sin^{-1}$ is the inverse function of $\sin$,

$$x=\sin y$$

Differentiating both sides with respect to $x$ gives

$$1 = {{d\sin y} \over {dx}} = { {d \sin } y \over {dy}} { {dy} \over {dx}} = y^{\prime} \cos y$$

Rearranging this with respect to $y^{\prime}$ yields

$$y^{\prime} = { {1} \over {\cos y} } = { {1} \over {\sqrt{1 - \sin ^{2} y}} } = { {1} \over {\sqrt{1 - x^2}} }$$

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