Porisma

Use principal inverse-trigonometric ranges

Derive and apply inverse-trigonometric derivatives with their range conditions.

Inverse trigonometric functions use principal ranges

Read the function domain and principal range before differentiating. Then check the stricter conditions needed by the displayed derivative formula; endpoint composites require their own analysis.

Here sin⁡−1\sin^{-1}, cos⁡−1\cos^{-1} and tan⁡−1\tan^{-1} mean arcsine, arccosine and arctangent. They are not 1/sin⁡1/\sin, 1/cos⁡1/\cos or 1/tan⁡1/\tan; the source explicitly distinguishes inverse functions from reciprocals.

Inverse function: sin⁡−1x\sin^{-1}x

  • Real domain: [−1,1][-1,1]
  • Principal range: [−π/2,π/2][-\pi/2,\pi/2]

Inverse function: cos⁡−1x\cos^{-1}x

  • Real domain: [−1,1][-1,1]
  • Principal range: [0,π][0,\pi]

Inverse function: tan⁡−1x\tan^{-1}x

  • Real domain: R\mathbb R
  • Principal range: (−π/2,π/2)(-\pi/2,\pi/2)

For a differentiable inner function u, the three source basic/general rows are:

Function: sin⁡−1u\sin^{-1}u

  • Basic derivative (u=x): 1/1−x21/\sqrt{1-x^2}
  • General derivative: u′/1−u2u^{\prime}/\sqrt{1-u^2}

Function: cos⁡−1u\cos^{-1}u

  • Basic derivative (u=x): −1/1−x2-1/\sqrt{1-x^2}
  • General derivative: −u′/1−u2-u^{\prime}/\sqrt{1-u^2}

Function: tan⁡−1u\tan^{-1}u

  • Basic derivative (u=x): 1/(1+x2)1/(1+x^2)
  • General derivative: u′/(1+u2)u^{\prime}/(1+u^2)

The source shades the three basic formulas as being in MF27. It prints −1≤x≤1-1\le x\le1 (and −1≤u≤1-1\le u\le1) beside the first two derivative rules. Those are the function domains: the displayed derivative quotients require ∣x∣<1|x|<1 or ∣u∣<1|u|<1 so their denominators are nonzero.

Endpoint composites need separate analysis; a constant inner function can, for example, give a constant endpoint value with derivative zero.

The arctangent rule has no corresponding real endpoint restriction.

Why arcsine uses the positive square root

The source proof sets y=sin⁡−1xy=\sin^{-1}x, so sin⁡y=x\sin y=x.

Implicit differentiation gives cos⁡y y′=1\cos y\,y^{\prime}=1 and y′=1/cos⁡yy^{\prime}=1/\cos y.

On the principal range [−π/2,π/2][-\pi/2,\pi/2], cos⁡y≥0\cos y\ge0, hence cos⁡y=1−sin⁡2y=1−x2\cos y=\sqrt{1-\sin^2y}=\sqrt{1-x^2}.

For −1<x<1-1<x<1, cosine is strictly positive, so division gives y′=1/1−x2y^{\prime}=1/\sqrt{1-x^2}.

The endpoints are excluded from this finite derivative formula.

Both copies point to Appendix page 20, Examples D and E, for the arccosine and arctangent proofs; both proofs are now included as Appendix Examples D and E below.

Try these examples

Open an example in chapter practice. Try it first, then read its solution. Use “Learn this step” to return to this unit.