Derivatives of Inverse Trigonometric Functions

Concept and Approach

Inverse trigonometric functions are continuous on their domains (accepted without proof). To differentiate them we combine the chain rule with implicit differentiation: if $y = f^{-1}(x)$ then $x = f(y)$, and differentiating both sides w.r.t. $x$ gives $1 = f'(y) \cdot \frac{dy}{dx}$, so $\frac{dy}{dx} = \frac{1}{f'(y)}$. The work lies in expressing this in terms of $x$ using trigonometric identities and the domain restrictions of the inverse functions.

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Derivative of $\sin^{-1} x$

Let $y = \sin^{-1} x$, so $x = \sin y$. Differentiating both sides with respect to $x$:

This is defined only when $\cos y \neq 0$, i.e. $y \neq \pm \frac{\pi}{2}$, so $x \neq \pm 1$; the derivative exists for $x \in (-1, 1)$. Since $\sin y = x$,

As $y = \sin^{-1} x$ lies in $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, $\cos y$ is positive, so $\cos y = \sqrt{1 - x^2}$. Therefore:

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Derivative of $\cos^{-1} x$

Let $y = \cos^{-1} x$, so $x = \cos y$. Differentiating with respect to $x$:

This exists when $\sin y \neq 0$, i.e. $y \neq 0, \pi$, so $x \neq \pm 1$ and the domain is $x \in (-1, 1)$. Since $\cos y = x$, $\sin^2 y = 1 - x^2$; and on the range $[0, \pi]$, $\sin y$ is non-negative, so $\sin y = \sqrt{1 - x^2}$. Therefore:

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Derivative of $\tan^{-1} x$

Let $y = \tan^{-1} x$, so $x = \tan y$. Differentiating with respect to $x$:

Using $\sec^2 y = 1 + \tan^2 y = 1 + x^2$:

which is defined for all real $x$, since $1 + x^2 > 0$.

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Derivative of $\cot^{-1} x$

Let $y = \cot^{-1} x$, so $x = \cot y$. Differentiating with respect to $x$:

Using $\csc^2 y = 1 + \cot^2 y = 1 + x^2$:

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Derivative of $\sec^{-1} x$

Let $y = \sec^{-1} x$, so $x = \sec y$. Differentiating with respect to $x$:

This exists when $\sec y \tan y \neq 0$; for the range $[0, \pi] \setminus \{\frac{\pi}{2}\}$ this holds for $|x| > 1$. Since $\sec y = x$, $\tan^2 y = \sec^2 y - 1 = x^2 - 1$, with the sign of $\tan y$ depending on the quadrant:

• If $y \in [0, \frac{\pi}{2})$, then $\tan y \geq 0$, so $\tan y = \sqrt{x^2 - 1}$.
• If $y \in (\frac{\pi}{2}, \pi]$, then $\tan y \leq 0$, so $\tan y = -\sqrt{x^2 - 1}$.

Both cases combine as:

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Derivative of $\csc^{-1} x$

Let $y = \csc^{-1} x$, so $x = \csc y$. Differentiating with respect to $x$:

This exists when $\csc y \cot y \neq 0$; for the range $[-\frac{\pi}{2}, \frac{\pi}{2}] \setminus \{0\}$ this holds for $|x| > 1$. Since $\csc y = x$, $\cot^2 y = \csc^2 y - 1 = x^2 - 1$, with:

• If $y \in (0, \frac{\pi}{2}]$, then $\cot y \geq 0$, so $\cot y = \sqrt{x^2 - 1}$.
• If $y \in [-\frac{\pi}{2}, 0)$, then $\cot y \leq 0$, so $\cot y = -\sqrt{x^2 - 1}$.

Both cases combine as:

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Summary Table of Derivatives …