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Chapter 5 · Continuity and Differentiability

Miscellaneous · Q6: Differentiate a cot⁡−1\cot^{-1} of roots of 1±sin⁡x1\pm\sin x

Miscellaneous ExerciseDerivatives of inverse trigonometric functions · Miscellaneous Exercise
The question

Differentiate w.r.t. xx the function

cot⁡−1[1+sin⁡x+1−sin⁡x1+sin⁡x−1−sin⁡x],\cot^{-1}\left[\frac{\sqrt{1+\sin x} + \sqrt{1-\sin x}}{\sqrt{1+\sin x} - \sqrt{1-\sin x}}\right], 0<x<π2.0 < x < \frac{\pi}{2}.
Exam answer

Let yy be the given function. For 0<x<π20 \lt x \lt \frac{\pi}{2}⁠, 0<x2<π40 \lt \frac x2 \lt \frac{\pi}{4}⁠, so cos⁡x2>sin⁡x2>0\cos\frac x2 \gt \sin\frac x2 \gt 0⁠.

Using 1=cos⁡2x2+sin⁡2x21 = \cos^2\frac x2 + \sin^2\frac x2 and sin⁡x=2sin⁡x2cos⁡x2\sin x = 2\sin\frac x2\cos\frac x2⁠,

1+sin⁡x=(cos⁡x2+sin⁡x2)21−sin⁡x=(cos⁡x2−sin⁡x2)2\begin{aligned} 1 + \sin x &= \Big(\cos\frac x2 + \sin\frac x2\Big)^2 \\ 1 - \sin x &= \Big(\cos\frac x2 - \sin\frac x2\Big)^2 \end{aligned}

Both brackets are positive, so 1±sin⁡x=cos⁡x2±sin⁡x2\sqrt{1 \pm \sin x} = \cos\frac x2 \pm \sin\frac x2⁠. Hence

y=cot⁡−1[2cos⁡x22sin⁡x2]=cot⁡−1(cot⁡x2)=x2[as x2∈(0,π)]\begin{aligned} y &= \cot^{-1}\Bigg[\frac{2\cos\frac x2}{2\sin\frac x2}\Bigg] \\ &= \cot^{-1}\Big(\cot\frac x2\Big) \\ &= \frac x2 \quad \Big[\text{as } \frac x2 \in (0, \pi)\Big] \end{aligned}

∴ dydx=12\therefore\ \dfrac{dy}{dx} = \dfrac12 (0<x<π2)\big(0 \lt x \lt \dfrac{\pi}{2}\big)

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

Differentiating this directly would be painful. Instead, simplify first. With half angles, 1±sin⁡x1 \pm \sin x are perfect squares: (cos⁡x2±sin⁡x2)2\left(\cos\frac x2 \pm \sin\frac x2\right)^2⁠. On the given interval both square roots can be taken without absolute-value trouble, the fraction collapses to cot⁡x2\cot\frac{x}{2}⁠, and since x2\frac x2 lies in the principal range (0,π)(0,\pi) of cot⁡−1\cot^{-1}⁠, the whole function is just x2\frac{x}{2}⁠.

Given, to find, and the plan
Given
y=cot⁡−1Fy = \cot^{-1} F⁠, where FF is the fraction in the square brackets, for 0<x<π20 < x < \frac{\pi}{2}⁠.
To find
dydx\dfrac{dy}{dx}⁠.
Plan
(1) Check the fraction is defined. (2) Write 1±sin⁡x1 \pm \sin x as squares. (3) Take the square roots, using the sign of each base on the interval. (4) Simplify FF to cot⁡x2\cot\frac x2⁠. (5) Use the range of cot⁡−1\cot^{-1} to get y=x2y = \frac x2⁠. (6) Differentiate.

Basics you need

1Principal value of cot⁡−1\cot^{-1}

cot⁡−1:R→(0,π)\cot^{-1}: \mathbf{R} \to (0,\pi)⁠, and cot⁡−1(cot⁡φ)=φ\cot^{-1}(\cot\varphi) = \varphi exactly when φ∈(0,π)\varphi \in (0,\pi) (Chapter 2, Sections 2.2 and 2.3). Basics: composite functions

2Half-angle identities

sin⁡x=2sin⁡x2cos⁡x2\sin x = 2\sin\frac x2\cos\frac x2 and 1=cos⁡2x2+sin⁡2x21 = \cos^2\frac x2 + \sin^2\frac x2 (Class XI). Basics: trigonometric identities

3Square root of a square

a2=∣a∣\sqrt{a^2} = |a|⁠: it equals aa only when a≥0a \ge 0⁠. This is where the interval matters.

4Constant multiple rule

ddx(cx)=c\frac{d}{dx}(cx) = c for a constant cc (Class XI, Limits and Derivatives). Basics: derivative rules

Solution

  1. The fraction is defined on (0,π2)\left(0, \frac{\pi}{2}\right)

    For 0<x<π20 < x < \frac{\pi}{2}⁠, 0<sin⁡x<10 < \sin x < 1⁠. So 1+sin⁡x>01 + \sin x > 0 and 1−sin⁡x>01 - \sin x > 0⁠: both square roots are real. Also 1+sin⁡x>1−sin⁡x1 + \sin x > 1 - \sin x⁠, so

    1+sin⁡x−1−sin⁡x>0,\sqrt{1+\sin x} - \sqrt{1-\sin x} > 0,

    and the denominator is never zero. Call the fraction FF⁠; then y=cot⁡−1Fy = \cot^{-1}F is defined.

    Before simplifying we make sure there is something to simplify: real roots and a non-zero denominator. The square root is an increasing function, so a bigger number has a bigger root.
    sin⁡x\sin x increases from 0 to 1 on [0,π2]\left[0,\frac{\pi}{2}\right] (Class XI). Basics: trigonometric identities

  2. Write 1±sin⁡x1 \pm \sin x as perfect squares

    Let c=cos⁡x2c = \cos\frac x2 and s=sin⁡x2s = \sin\frac x2⁠. Then 1=c2+s21 = c^2 + s^2 and sin⁡x=2sc\sin x = 2sc⁠, so

    1+sin⁡x=c2+s2+2cs=(c+s)2,1−sin⁡x=c2+s2−2cs=(c−s)2.\begin{aligned} 1 + \sin x &= c^2 + s^2 + 2cs = (c + s)^2, \\ 1 - \sin x &= c^2 + s^2 - 2cs = (c - s)^2. \end{aligned}
    This is the step that removes the square roots. Writing 1 as cos⁡2+sin⁡2\cos^2 + \sin^2 of the half angle and sin⁡x\sin x as a double angle produces the expansions of (c±s)2(c \pm s)^2⁠.
    sin⁡2A=2sin⁡Acos⁡A\sin 2A = 2\sin A\cos A with A=x2A = \frac x2⁠; cos⁡2A+sin⁡2A=1\cos^2 A + \sin^2 A = 1 (Class XI). Basics: trigonometric identities

  3. Take the square roots, checking signs

    Since 0<x<π20 < x < \frac{\pi}{2}⁠, the half angle satisfies 0<x2<π40 < \frac{x}{2} < \frac{\pi}{4}⁠. On this range s>0s > 0⁠, c>0c > 0⁠, and c>sc > s (because tan⁡x2<1\tan\frac x2 < 1⁠). So both c+sc + s and c−sc - s are positive, and

    1+sin⁡x=∣c+s∣=c+s,1−sin⁡x=∣c−s∣=c−s.\begin{aligned} \sqrt{1+\sin x} &= |c + s| = c + s, \\ \sqrt{1-\sin x} &= |c - s| = c - s. \end{aligned}
    a2=∣a∣\sqrt{a^2} = |a|⁠, not aa⁠. For c+sc + s there is no issue, but c−sc - s is positive only while x2<π4\frac{x}{2} < \frac{\pi}{4}⁠. The given interval x<π2x < \frac{\pi}{2} is exactly what guarantees this. If x2\frac x2 were past π4\frac{\pi}{4}⁠, the root would be s−cs - c and the whole answer would change (see Common mistakes).
    tan⁡A<1\tan A < 1 for 0<A<π40 < A < \frac{\pi}{4}⁠, so sin⁡A<cos⁡A\sin A < \cos A there (Class XI). Basics: trigonometric identities

  4. Simplify the fraction to cot⁡x2\cot\frac{x}{2}

    F=(c+s)+(c−s)(c+s)−(c−s)=2c2s=cos⁡x2sin⁡x2=cot⁡x2.\begin{aligned} F &= \frac{(c + s) + (c - s)}{(c + s) - (c - s)} \\ &= \frac{2c}{2s} = \frac{\cos\frac x2}{\sin\frac x2} = \cot\frac{x}{2}. \end{aligned}
    The ss terms cancel on top and the cc terms cancel underneath. Dividing by 2s2s is allowed because s=sin⁡x2>0s = \sin\frac{x}{2} > 0 on the interval.
    cot⁡A=cos⁡Asin⁡A\cot A = \frac{\cos A}{\sin A} for sin⁡A≠0\sin A \ne 0⁠.

  5. Use the principal range of cot⁡−1\cot^{-1}

    Now y=cot⁡−1(cot⁡x2)y = \cot^{-1}\left(\cot\frac{x}{2}\right)⁠. The angle x2\frac{x}{2} lies in (0,π4)\left(0, \frac{\pi}{4}\right)⁠, which is inside (0,π)(0,\pi)⁠, the range of cot⁡−1\cot^{-1}⁠. Therefore

    y=x2,0<x<π2.y = \frac{x}{2}, \quad 0 < x < \frac{\pi}{2}.
    cot⁡−1\cot^{-1} returns the unique angle in (0,π)(0,\pi) with the given cotangent. Since x2\frac x2 is such an angle, it is the one returned. This is the same identity as Chapter 2, Miscellaneous Exercise Q9 (stated there for x∈(0,π4)x \in \left(0,\frac{\pi}{4}\right)⁠); the proof above shows it holds on the larger interval (0,π2)\left(0,\frac{\pi}{2}\right) used here.
    cot⁡−1(cot⁡φ)=φ\cot^{-1}(\cot\varphi) = \varphi for φ∈(0,π)\varphi \in (0,\pi) (Chapter 2, Section 2.3, "similar results follow"). Basics: composite functions

  6. Differentiate

    dydx=ddx(x2)=12.\frac{dy}{dx} = \frac{d}{dx}\left(\frac{x}{2}\right) = \frac12.
    On the whole interval the function is the straight line y=x2y = \frac x2⁠, whose slope is 12\frac12 everywhere.
    Constant multiple rule (Class XI); ddxx=1\frac{d}{dx}x = 1 (Table 5.3). Basics: derivative rules
0 π π/2 π/4 x/2 lies in (0, π/4) x lies in (0, π/2) range of cot⁻¹: (0, π) angle π/4: cos = sin; below it, cos > sin
Angles measured from the positive horizontal axis. The half angle x2\frac x2 (gold) stays below π4\frac{\pi}{4}⁠, so cos⁡x2>sin⁡x2>0\cos\frac x2 > \sin\frac x2 > 0 and both square roots in Step 3 are taken without a sign change. The gold arc also lies inside the principal range (0,π)(0,\pi) of cot⁡−1\cot^{-1} (violet), so cot⁡−1(cot⁡x2)=x2\cot^{-1}\left(\cot\frac x2\right) = \frac x2⁠. Open circles mark excluded end points.
Result

For 0<x<π20 < x < \frac{\pi}{2}⁠, the function equals x2\frac{x}{2}⁠, so

dydx=12.\frac{dy}{dx} = \frac{1}{2}.

(Textbook answer: 12\frac12⁠, the same.)

Check it with numbers

At x=1x = 1 (inside (0,π2)\left(0, \frac{\pi}{2}\right)⁠, since π2≈1.5708\frac{\pi}{2} \approx 1.5708⁠): sin⁡1≈0.8414710\sin 1 \approx 0.8414710⁠, 1.8414710≈1.3570081\sqrt{1.8414710} \approx 1.3570081 and 0.1585290≈0.3981570\sqrt{0.1585290} \approx 0.3981570⁠. These equal cos⁡0.5+sin⁡0.5\cos 0.5 + \sin 0.5 and cos⁡0.5−sin⁡0.5\cos 0.5 - \sin 0.5⁠, as Step 3 says.

F=1.75516510.9588511≈1.8304877F = \frac{1.7551651}{0.9588511} \approx 1.8304877⁠, which is cot⁡0.5\cot 0.5⁠. Then y=cot⁡−1(1.8304877)y = \cot^{-1}(1.8304877)⁠, which is 0.5000000=12⋅10.5000000 = \frac12\cdot 1⁠. ✓

With h=0.0001h = 0.0001⁠: y(1.0001)=0.50005y(1.0001) = 0.50005 and y(0.9999)=0.49995y(0.9999) = 0.49995⁠, so the difference quotient is 0.50.5⁠. ✓

Common mistakes
  1. Writing 1−sin⁡x=cos⁡x2−sin⁡x2\sqrt{1-\sin x} = \cos\frac x2 - \sin\frac x2 without checking the sign. This is true only for x2≤π4\frac x2 \le \frac{\pi}{4}⁠. For example, for π2<x<π\frac{\pi}{2} < x < \pi the root is sin⁡x2−cos⁡x2\sin\frac x2 - \cos\frac x2⁠, the fraction becomes tan⁡x2=cot⁡(π2−x2)\tan\frac x2 = \cot\left(\frac{\pi}{2} - \frac x2\right)⁠, and y=π2−x2y = \frac{\pi}{2} - \frac{x}{2} with derivative −12-\frac12⁠. At x=2x = 2 the function equals π2−1≈0.5708\frac{\pi}{2} - 1 \approx 0.5708⁠, not 1.
  2. Differentiating the fraction directly. It works in principle but is long and error-prone; simplify first.
  3. Skipping the range check for cot⁡−1\cot^{-1}⁠. cot⁡−1(cot⁡φ)=φ\cot^{-1}(\cot\varphi) = \varphi needs φ∈(0,π)\varphi \in (0,\pi)⁠; it is not automatic.
What to take away
  1. 1±sin⁡x=(cos⁡x2±sin⁡x2)21 \pm \sin x = \left(\cos\frac x2 \pm \sin\frac x2\right)^2 is the standard way to remove 1±sin⁡x\sqrt{1 \pm \sin x}⁠.
  2. Every square root and every inverse-trig cancellation must be checked on the stated interval; here both checks come down to 0<x2<π40 < \frac x2 < \frac{\pi}{4}⁠.

Every fact used, and where it comes from

FactUsed inSource
cot⁡−1:R→(0,π)\cot^{-1}: \mathbf{R} \to (0,\pi)Step 5Chapter 2, Section 2.2
cot⁡−1(cot⁡φ)=φ\cot^{-1}(\cot\varphi) = \varphi for φ∈(0,π)\varphi \in (0,\pi)Step 5Chapter 2, Section 2.3
The same simplification (on a smaller interval)Steps 2–5Chapter 2, Miscellaneous Exercise Q9
sin⁡2A=2sin⁡Acos⁡A\sin 2A = 2\sin A\cos A⁠; cos⁡2A+sin⁡2A=1\cos^2A + \sin^2A = 1Step 2Class XI, Trigonometric Functions
a2=∣a∣\sqrt{a^2} = |a|⁠; signs of sin⁡\sin⁠, cos⁡\cos on (0,π4)\left(0, \frac{\pi}{4}\right)Steps 1, 3Class XI
ddxx=1\frac{d}{dx}x = 1⁠, constant multiplesStep 6Table 5.3 and rules of Section 5.3 (Basics: derivative rules); constant multiples, Class XI
AnswerResultAnswers, Miscellaneous Exercise on Chapter 5, Q6
Numerical values, figure coordinatesCheck, figureRecomputed in the site's verification script
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