An inverse-trigonometric identity and its endpoints

(i) Show that ddx(cos⁡−12x1+x2)=−21+x2\frac{d}{dx}\left(\cos^{-1}\frac{2x}{1+x^2}\right)=\frac{-2}{1+x^2} for −1≤x≤1-1\le x\le1. Explain, in your working, how you use the condition −1≤x≤1-1\le x\le1 to conclude your answer. marks

(ii) Hence, determine the constants AA and BB satisfying the equation cos⁡−12x1+x2+Atan⁡−1x=B\cos^{-1}\frac{2x}{1+x^2}+A\tan^{-1}x=B, for −1≤x≤1-1\le x\le1. marks

[2009/YJC/Prelim/I/5(a)]

Total marks: 6

Solution

(i)

Chain rule with inverse trigonometry

Show that ddx(cos⁡−12x1+x2)=−21+x2\dfrac d{dx}\left(\cos^{-1}\dfrac{2x}{1+x^2}\right)=\dfrac{-2}{1+x^2} for −1≤x≤1-1\le x\le1, and explain how you use the condition −1≤x≤1-1\le x\le1.

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Hint

  1. Use ddxcos⁡−1u=−11−u2 u′\frac d{dx}\cos^{-1}u=-\frac1{\sqrt{1-u^2}}\,u' with u=2x1+x2u=\frac{2x}{1+x^2}.

  2. Differentiate uu by the quotient rule.

  3. Simplify the square root: it becomes ∣1−x2∣|1-x^2|.

Answer · (i)

d[arccos⁡(2x/(1+x2))]/dx=−2/(1+x2)d[\arccos(2x/(1+x^2))]/dx=-2/(1+x^2).

For −1<x<1. Inward one-sided derivatives at ±1 are −1; unrestricted two-sided derivatives do not exist there.

Explore the idea

Open interactive exploration

arccos(2x/(1 + x²)) is a falling S-curve

The gradient graph is −2/(1 + x²): always negative, steepest at x = 0 where it equals −2.

Position along the curvex = 0
Try an example
f(x)
1.5708
Gradient f′(x)
-2
negative

Inside (−1,1) the gradient is −2/(1+x²)<0. At x=0 it is −2.

Selected values are within the original drawing windows.

  • Y = cos⁻¹(2x/(1 + x²))
  • Tangent
  • Gradient −2/(1 + x²)
  • Join P to Q; the dashed legs show Δx and Δy
Key idea

On [−1,1][-1,1] the curve falls from π\pi to 00, with gradient −21+x2-\frac2{1+x^2}.

Ideas to try
At x=0x=0 the curve is at height π2\frac\pi2. What is its gradient there?