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What is the derivative of the inverse trig functions

The derivative of arcsin⁡x\arcsin x is 11−x2\frac{1}{\sqrt{1-x^{2}}}, of arccos⁡x\arccos x is −11−x2-\frac{1}{\sqrt{1-x^{2}}}, of arctan⁡x\arctan x is 11+x2\frac{1}{1+x^{2}}, of \(\arccot x\) is −11+x2-\frac{1}{1+x^{2}}, of \(\arcsec x\) is 1∣x∣x2−1\frac{1}{|x|\sqrt{x^{2}-1}}, and of \(\arccsc x\) is −1∣x∣x2−1-\frac{1}{|x|\sqrt{x^{2}-1}}.

Calculus · Derivatives

Inverse trigonometric functions are the angles whose trigonometric values equal a given number. Their derivatives follow from implicit differentiation and the Pythagorean identity.

Standard derivative formulas

Key results

  • ddxarcsin⁡x=11−x2\frac{d}{dx}\arcsin x = \frac{1}{\sqrt{1-x^{2}}}
  • ddxarccos⁡x=−11−x2\frac{d}{dx}\arccos x = -\frac{1}{\sqrt{1-x^{2}}}
  • ddxarctan⁡x=11+x2\frac{d}{dx}\arctan x = \frac{1}{1+x^{2}}
  • \(\frac{d}{dx}\arccot x = -\frac{1}{1+x^{2}}\)
  • \(\frac{d}{dx}\arcsec x = \frac{1}{|x|\sqrt{x^{2}-1}}\)
  • \(\frac{d}{dx}\arccsc x = -\frac{1}{|x|\sqrt{x^{2}-1}}\)

Deriving arcsin⁡x\arcsin x as an example

  1. 1Set y=arcsin⁡xy=\arcsin x so sin⁡y=x\sin y = x.
  2. 2Differentiate both sides: cos⁡y dy/dx=1\cos y\,dy/dx = 1.
  3. 3Use cos⁡y=1−sin⁡2y=1−x2\cos y = \sqrt{1-\sin^{2}y}=\sqrt{1-x^{2}}.
  4. 4Solve for dy/dxdy/dx: dy/dx=1/1−x2dy/dx = 1/\sqrt{1-x^{2}}.

Summary of all six inverse trig derivatives

FunctionDerivative
arcsin⁡\arcsin x11−x2\frac{1}{\sqrt{1-x^{2}}}
arccos⁡\arccos x-11−x2\frac{1}{\sqrt{1-x^{2}}}
arctan⁡\arctan x11+x2\frac{1}{1+x^{2}}
\arccot x-11+x2\frac{1}{1+x^{2}}
\arcsec x1∣x∣x2−1\frac{1}{|x|\sqrt{x^{2}-1}}
\arccsc x-1∣x∣x2−1\frac{1}{|x|\sqrt{x^{2}-1}}

Check yourself

What is the derivative of arccos⁡x\arccos x?

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