Derivatives of Inverse Trigonometric Functions

This lesson derives the derivatives of all six inverse trigonometric functions. The method is the same each time: set $y$ equal to the inverse function, rewrite it in direct form, differentiate implicitly, and then express the result in terms of $x$.

Derivative of inverse sine

Let $y = \sin^{-1} x$, so that $x = \sin y$. This is an implicit relation, so differentiate with respect to $y$:

$\frac{dx}{dy} = \cos y.$

Taking the reciprocal gives

$\frac{dy}{dx} = \frac{1}{\cos y}.$

Since $\cos y = \sqrt{1 - \sin^2 y}$ and $\sin y = x$, we get $\cos y = \sqrt{1 - x^2}$. Therefore

$\frac{d}{dx}\,\sin^{-1} x = \frac{1}{\sqrt{1 - x^2}}.$

Derivative of inverse cosine

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

$\frac{dx}{dy} = -\sin y, \qquad \frac{dy}{dx} = -\frac{1}{\sin y}.$

With $\sin y = \sqrt{1 - \cos^2 y}$ and $\cos y = x$, this becomes

$\frac{d}{dx}\,\cos^{-1} x = -\frac{1}{\sqrt{1 - x^2}}.$

Derivative of inverse tangent

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

$\frac{dx}{dy} = \sec^2 y, \qquad \frac{dy}{dx} = \frac{1}{\sec^2 y} = \frac{1}{1 + \tan^2 y}.$

Since $\tan y = x$,

$\frac{d}{dx}\,\tan^{-1} x = \frac{1}{1 + x^2}.$

Derivative of inverse cotangent

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

$\frac{dx}{dy} = -\csc^2 y, \qquad \frac{dy}{dx} = -\frac{1}{\csc^2 y} = -\frac{1}{1 + \cot^2 y}.$

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

$\frac{d}{dx}\,\cot^{-1} x = -\frac{1}{1 + x^2}.$

Derivative of inverse secant

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

$\frac{dx}{dy} = \sec y \tan y, \qquad \frac{dy}{dx} = \frac{1}{\sec y \tan y}.$

Since $\tan y = \sqrt{\sec^2 y - 1}$ and $\sec y = x$,

$\frac{d}{dx}\,\sec^{-1} x = \frac{1}{x\,\sqrt{x^2 - 1}}.$

Derivative of inverse cosecant

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

$\frac{dx}{dy} = -\csc y \cot y, \qquad \frac{dy}{dx} = -\frac{1}{\csc y \cot y}.$

Since $\cot y = \sqrt{\csc^2 y - 1}$ and $\csc y = x$,

$\frac{d}{dx}\,\csc^{-1} x = -\frac{1}{x\,\sqrt{x^2 - 1}}.$

Key takeaways

• Each formula comes from the same routine: set $y$ to the inverse function, write the direct relation, differentiate implicitly, then take the reciprocal.
• The sine, tangent, and secant cases are positive; their co-function partners (cosine, cotangent, cosecant) are the negatives of the same expressions.
• The final forms are $\dfrac{1}{\sqrt{1 - x^2}}$, $\dfrac{1}{1 + x^2}$, and $\dfrac{1}{x\sqrt{x^2 - 1}}$, with a minus sign for the co-functions.