📂Functions

Derivatives of Inverse Trigonometric Functions

Formulas

Inverse trigonometric functions’ derivatives are as follows.

$$\begin{align*} \dfrac{d}{dx} \sin ^{-1} x &= \dfrac{1}{\sqrt{1-x^2}} \qquad & \dfrac{d}{dx} \csc ^{-1} x &= -\dfrac{1}{x\sqrt{x^2-1}} \\ \dfrac{d}{dx} \cos ^{-1} x &= -\dfrac{1}{\sqrt{1-x^2}} \qquad & \dfrac{d}{dx} \sec ^{-1} x &= \dfrac{1}{x\sqrt{x^2-1}} \\ \dfrac{d}{dx} \tan ^{-1} x &= \dfrac{1}{1+x^2} \qquad & \dfrac{d}{dx} \cot ^{-1} x &= -\dfrac{1}{1+x^2} \end{align*}$$

Proof

Differentiation of trigonometric functions

$$\begin{align*} \dfrac{d}{dx} \sin x &= \cos x \qquad & \dfrac{d}{dx} \csc x &= -\csc x \cot x \\[1em] \dfrac{d}{dx} \cos x &= - \sin x \qquad & \dfrac{d}{dx} \sec x &= \sec x \tan x \\[1em] \dfrac{d}{dx} \tan x &= \sec^{2} x \qquad & \dfrac{d}{dx} \cot x &= -\csc^{2} x \end{align*}$$

$(\sin^{-1})^{\prime}$, $(\cos^{-1})^{\prime}$

Let’s denote it as $y = \sin^{-1} x$. Then $\sin y = x$ and the range of $y$ is $- \dfrac{\pi}{2} \le y \le \dfrac{\pi}{2}$. Differentiating with respect to $x$, by the chain rule, we get the following.

$$\dfrac{d}{dx}\sin y = (x)^{\prime} \implies \dfrac{d}{dy}\sin y \cdot \dfrac{y}{dx} = 1 \implies \cos y \cdot \dfrac{dy}{dx} = 1$$

Therefore, we obtain the following.

$$\dfrac{dy}{dx} = \dfrac{1}{\cos y} = \dfrac{1}{\sqrt{1 - \sin^{2} y}}$$

But since it was $\sin y = x$,

$$\dfrac{dy}{dx} = \dfrac{1}{\sqrt{1 - x^{2}}}$$

$\dfrac{d}{dx} \cos ^{-1} x$ is obtained in the same way.

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$(\tan^{-1})^{\prime}$

Let’s denote it as $y = \tan^{-1} x$. Then $\tan y = x$ and the range of $y$ is $-\dfrac{\pi}{2} \lt y \lt \dfrac{\pi}{2}$. Differentiating with respect to $x$, by the chain rule, we get the following.

$$\dfrac{d}{dx} \tan y = (x)^{\prime} \implies \dfrac{d}{dy} \tan y \cdot \dfrac{y}{dx} = 1 \implies \sec^{2} y \cdot \dfrac{dy}{dx} = 1$$

Therefore, we obtain the following.

$$\dfrac{dy}{dx} = \dfrac{1}{\sec^{2} y} = \dfrac{1}{1 + \tan^{2} y}$$

But since it was $\tan y = x$,

$$\dfrac{dy}{dx} = \dfrac{1}{1 + x^{2}}$$

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$(\csc^{-1})^{\prime}$, $(\sec^{-1})^{\prime}$

Let’s denote it as $y = \csc^{-1} x$. Then $\csc y = x$ and differentiating both sides with respect to $x$, by the chain rule, we get the following.

$$\dfrac{d}{dx} \csc y = (x)^{\prime} \implies \dfrac{d}{dy} \csc y \dfrac{dy}{dx} = 1 \implies -\csc y \cot y \dfrac{dy}{dx} = 1$$

Therefore, we obtain the following.

$$\dfrac{dy}{dx} = -\dfrac{1}{\csc y \cot y} = - \sin y \tan y$$

But since it was $\csc y = x$ and $\sin y = \dfrac{1}{x}$, it results in $\cos y = \sqrt{1 - \dfrac{1}{x^{2}}} = \dfrac{1}{x}\sqrt{x^{2} - 1}$. Therefore,

$$\dfrac{dy}{dx} = -\dfrac{1}{x\sqrt{x^{2} - 1}}$$

$(\sec ^{-1} x)^{\prime}$ is also obtained in the same way.

$(\cot^{-1})^{\prime}$

Let’s denote it as $y = \cot^{-1} x$. Then $\cot y = x$ and differentiating both sides with respect to $x$, by the chain rule, we get the following.

$$\dfrac{d}{dx} \cot y = (x)^{\prime} \implies \dfrac{d}{dy} \cot y \dfrac{dy}{dx} = 1 \implies -\csc^{2} y \dfrac{dy}{dx} = 1$$

Therefore, we obtain the following.

$$\dfrac{dy}{dx} = -\dfrac{1}{\csc^{2} y} = - \sin^{2}y$$

But since it was stated as $\cot y = x$,

$$\dfrac{\cos y}{\sin y} = x \implies \dfrac{\cos^{2} y}{\sin^{2}y} = x^{2} \implies \dfrac{1 - \sin^{2}y}{\sin^{2}y} = x^{2}$$

Organizing with respect to $\sin^{2}y$, we get $\sin^{2}y = \dfrac{1}{1 + x^{2}}$. Therefore,

$$\dfrac{dy}{dx} = - \dfrac{1}{1 + x^{2}}$$

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