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Inverse trig derivatives

Derivative of arcsec x: Formula, Proof and Examples

The derivative of arcsec x is 1/(|x|√(x² − 1)). See where the absolute value comes from, a proof via arccos(1/x), and chain rule examples.

Inverse trig derivatives Updated 2 min read

At a glance

\frac{d}{dx}\,\operatorname{arcsec} x = \frac{1}{\lvert x\rvert\sqrt{x^2 - 1}}
Derivative
\dfrac{1}{\lvert x\rvert\sqrt{x^2 - 1}} for \lvert x\rvert > 1
Chain rule form
\dfrac{d}{dx}\operatorname{arcsec} u = \dfrac{u'}{\lvert u\rvert\sqrt{u^2 - 1}}
Domain of arcsec x
x \le -1 or x \ge 1
Range of arcsec x
\left[0, \frac{\pi}{2}\right) \cup \left(\frac{\pi}{2}, \pi\right]
Arccsc derivative
-\dfrac{1}{\lvert x\rvert\sqrt{x^2 - 1}}
On this page
  1. Graph of arcsec x
  2. Proof using arccos
  3. Chain rule version and examples
  4. The arccsc derivative
  5. Common mistakes

The derivative of \operatorname{arcsec} x is \dfrac{1}{\lvert x\rvert\sqrt{x^2 - 1}}. It is defined for \lvert x\rvert > 1, the same values where secant can output a number.

\frac{d}{dx}\,\operatorname{arcsec} x = \frac{1}{\lvert x\rvert\sqrt{x^2 - 1}}, \qquad \lvert x\rvert > 1

For x > 1, where most problems live, this is just \frac{1}{x\sqrt{x^2 - 1}}.

Graph of arcsec x

Graph of y = arcsec x, drawn as arccos(1/x)
Both pieces of the inverse secant rise. The right piece starts at (1, 0) and the left piece ends at (-1, \pi). Both approach the line y = \frac{\pi}{2}.

Both branches go upward from left to right, so the derivative is positive everywhere. That is exactly why the formula has \lvert x\rvert in it: without the absolute value, the formula would give negative slopes for x < -1, which does not match the picture.

Proof using arccos

The usual range of arcsec makes this identity true:

\operatorname{arcsec} x = \arccos\frac{1}{x}

Differentiate with the chain rule and the derivative of arccos x:

\frac{d}{dx}\arccos\frac{1}{x} = -\frac{1}{\sqrt{1 - \frac{1}{x^2}}}\cdot\left(-\frac{1}{x^2}\right) = \frac{1}{x^2\sqrt{1 - \frac{1}{x^2}}}

Now simplify the root: \sqrt{1 - \frac{1}{x^2}} = \frac{\sqrt{x^2 - 1}}{\sqrt{x^2}} = \frac{\sqrt{x^2 - 1}}{\lvert x\rvert}. So

\frac{d}{dx}\operatorname{arcsec} x = \frac{\lvert x\rvert}{x^2\sqrt{x^2 - 1}} = \frac{1}{\lvert x\rvert\sqrt{x^2 - 1}}

using x^2 = \lvert x\rvert^2. The absolute value appears naturally from \sqrt{x^2} = \lvert x\rvert.

Chain rule version and examples

\frac{d}{dx}\operatorname{arcsec} u = \frac{u'}{\lvert u\rvert\sqrt{u^2 - 1}}

Example 1. \frac{d}{dx}\operatorname{arcsec}(x^2) = \frac{2x}{x^2\sqrt{x^4 - 1}} = \frac{2}{x\sqrt{x^4 - 1}} for \lvert x\rvert > 1. Here \lvert x^2\rvert = x^2, so no absolute value is left.

Example 2. For x > 0, \frac{d}{dx}\operatorname{arcsec}(e^x) = \frac{e^x}{e^x\sqrt{e^{2x} - 1}} = \frac{1}{\sqrt{e^{2x} - 1}}.

Example 3: slope at a point. At x = 2, the slope of \operatorname{arcsec} x is \frac{1}{2\sqrt{3}} \approx 0.289. Since \operatorname{arcsec} 2 = \frac{\pi}{3}, the tangent line is y = \frac{\pi}{3} + \frac{1}{2\sqrt{3}}(x - 2).

The arccsc derivative

Since \operatorname{arcsec} x + \operatorname{arccsc} x = \frac{\pi}{2}, the derivatives are opposites:

\frac{d}{dx}\operatorname{arccsc} x = -\frac{1}{\lvert x\rvert\sqrt{x^2 - 1}}

Common mistakes

  1. Dropping the absolute value and then getting negative slopes for x < -1.
  2. Mixing textbook conventions. Check which range your course uses for arcsec before copying a formula.
  3. Confusing \sec^{-1} x with \cos x. \sec^{-1} is the inverse of secant, while \frac{1}{\sec x} = \cos x.
  4. Using it for -1 < x < 1. No angle has a secant between -1 and 1, so arcsec and its derivative do not exist there.

The formula also gives an integral: \int \frac{dx}{x\sqrt{x^2 - 1}} = \operatorname{arcsec}\lvert x\rvert + C. For more, see inverse trig derivative problems and the reciprocal trig functions.

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Frequently asked questions

What is the derivative of arcsec x?
The derivative of \operatorname{arcsec} x is \frac{1}{\lvert x\rvert\sqrt{x^2 - 1}} for \lvert x\rvert > 1. For x > 1 this is simply \frac{1}{x\sqrt{x^2 - 1}}.

Why do some books write x√(x² − 1) without the absolute value?
Those books pick a different range for arcsec, \left[0, \frac{\pi}{2}\right) \cup \left[\pi, \frac{3\pi}{2}\right). With that choice the formula is \frac{1}{x\sqrt{x^2 - 1}}. With the more common range [0, \pi] you need \lvert x\rvert.

What is the derivative of arccsc x?
It is -\frac{1}{\lvert x\rvert\sqrt{x^2 - 1}}, the negative of the arcsec derivative, because \operatorname{arcsec} x + \operatorname{arccsc} x = \frac{\pi}{2}.

What is the derivative of arcsec(2x)?
By the chain rule it is \frac{2}{\lvert 2x\rvert\sqrt{4x^2 - 1}} = \frac{1}{\lvert x\rvert\sqrt{4x^2 - 1}}, valid for \lvert x\rvert > \frac{1}{2}.

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Related derivatives

Derivative of arcsin x \frac{d}{dx}\,\arcsin x = \frac{1}{\sqrt{1 - x^2}} The derivative of arcsin x is 1/√(1 − x²). See the proof from sin y = x, the chain rule form for arcsin(u), a graph and worked examples. Derivative of arccos x \frac{d}{dx}\,\arccos x = -\frac{1}{\sqrt{1 - x^2}} The derivative of arccos x is −1/√(1 − x²). See the proof, why it is the negative of the arcsin derivative, chain rule examples and a graph. Derivative of arctan x \frac{d}{dx}\,\arctan x = \frac{1}{1 + x^2} The derivative of arctan x is 1/(1 + x²). See the proof from tan y = x, chain rule examples like arctan(2x) and arctan(x²), and a graph.