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Chapter 5 · Continuity and Differentiability

Exercise 5.3 · Q15: dydx\frac{dy}{dx} if y=sec⁡−112x2−1y=\sec^{-1}\frac{1}{2x^2-1}

Exercise 5.3Derivatives of inverse trigonometric functions · Section 5.3.3
The question

Find dydx\dfrac{dy}{dx} in the following:

y=sec⁡−1(12x2−1),0<x<12y = \sec^{-1}\left(\frac{1}{2x^2 - 1}\right), \quad 0 < x < \frac{1}{\sqrt2}
Exam answer

Put x=cos⁡θx = \cos\theta⁠. As 0<x<120 < x < \frac{1}{\sqrt2}⁠, π4<θ<π2\frac{\pi}{4} < \theta < \frac{\pi}{2}⁠, so π2<2θ<π\frac{\pi}{2} < 2\theta < \pi⁠.

y=sec⁡−1(12cos⁡2θ−1)=sec⁡−1(1cos⁡2θ)=sec⁡−1(sec⁡2θ)=2θ=2cos⁡−1x\begin{aligned} y &= \sec^{-1}\left(\frac{1}{2\cos^2\theta - 1}\right) \\ &= \sec^{-1}\left(\frac{1}{\cos 2\theta}\right) \\ &= \sec^{-1}(\sec 2\theta) = 2\theta \\ &= 2\cos^{-1}x \end{aligned}

∴ dydx=−21−x2\therefore\ \dfrac{dy}{dx} = -\dfrac{2}{\sqrt{1 - x^2}} (0<x<12)\left(0 < x < \dfrac{1}{\sqrt2}\right)⁠.

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

The one idea this problem is built on

The book's table of derivatives covers sin⁡−1\sin^{-1}⁠, cos⁡−1\cos^{-1} and tan⁡−1\tan^{-1} but not sec⁡−1\sec^{-1}⁠, so we simplify first. With x=cos⁡θx = \cos\theta⁠, 2x2−1=cos⁡2θ2x^2 - 1 = \cos 2\theta⁠, so the inside is 1cos⁡2θ=sec⁡2θ\frac{1}{\cos 2\theta} = \sec 2\theta⁠, and y=sec⁡−1(sec⁡2θ)y = \sec^{-1}(\sec 2\theta)⁠. This equals 2θ2\theta when 2θ2\theta lies in [0,π][0, \pi] with 2θ≠π22\theta \ne \frac{\pi}{2}, the principal range of sec⁡−1\sec^{-1}⁠. For 0<x<120 < x < \frac{1}{\sqrt2} we get 2θ∈(π2,π)2\theta \in \left(\frac{\pi}{2}, \pi\right)⁠, so y=2cos⁡−1xy = 2\cos^{-1}x⁠.

Given, to find, and the plan
Given
y=sec⁡−112x2−1y = \sec^{-1}\frac{1}{2x^2 - 1} for 0<x<120 < x < \frac{1}{\sqrt2}⁠.
To find
dydx\dfrac{dy}{dx} on that interval.
Plan
(1) Put x=cos⁡θx = \cos\theta and locate θ\theta⁠. (2) Show the inside is sec⁡2θ\sec 2\theta⁠, and that it is in the domain of sec⁡−1\sec^{-1}⁠. (3) Check 2θ2\theta is in the principal range and simplify yy⁠. (4) Differentiate. (5) Confirm by a second route through cos⁡−1\cos^{-1}⁠.

Basics you need

1Principal values of sec⁡−1\sec^{-1} and cos⁡−1\cos^{-1}

sec⁡−1:R−(−1,1)→[0,π]−{π2}\sec^{-1}: \mathbf{R} - (-1, 1) \to [0, \pi] - \left\{\frac{\pi}{2}\right\} and cos⁡−1:[−1,1]→[0,π]\cos^{-1}: [-1, 1] \to [0, \pi]⁠, decreasing (Chapter 2, Section 2.2). sec⁡−1(sec⁡ϕ)=ϕ\sec^{-1}(\sec\phi) = \phi for ϕ\phi in that range (Section 2.3). Basics: composite functions

2Double angle for cosine

cos⁡2θ=2cos⁡2θ−1\cos 2\theta = 2\cos^2\theta - 1⁠, and sec⁡ϕ=1cos⁡ϕ\sec\phi = \frac{1}{\cos\phi} (Class XI). Basics: trigonometric identities

3Special values

cos⁡π4=12\cos\frac{\pi}{4} = \frac{1}{\sqrt2}⁠, cos⁡π2=0\cos\frac{\pi}{2} = 0⁠; so cos⁡−112=π4\cos^{-1}\frac{1}{\sqrt2} = \frac{\pi}{4}⁠, cos⁡−10=π2\cos^{-1}0 = \frac{\pi}{2}⁠.

4Derivative of cos⁡−1x\cos^{-1}x

ddxcos⁡−1x=−11−x2\frac{d}{dx}\cos^{-1}x = \frac{-1}{\sqrt{1 - x^2}} for −1<x<1-1 < x < 1 (table after Example 24). Lesson: derivatives of inverse trig functions

Solution

  1. Substitute x=cos⁡θx = \cos\theta and locate θ\theta

    Let θ=cos⁡−1x\theta = \cos^{-1}x⁠. This is defined since 0<x<10 < x < 1⁠, and cos⁡θ=x\cos\theta = x⁠. Because cos⁡−1\cos^{-1} is decreasing, 0<x<120 < x < \frac{1}{\sqrt2} gives

    π4<θ<π2,soπ2<2θ<π.\frac{\pi}{4} < \theta < \frac{\pi}{2}, \quad\text{so}\quad \frac{\pi}{2} < 2\theta < \pi.
    A decreasing function reverses inequalities: the smaller end x=0x = 0 goes to the larger angle π2\frac{\pi}{2}⁠, and x=12x = \frac{1}{\sqrt2} goes to π4\frac{\pi}{4}⁠.
    cos⁡−1\cos^{-1} is decreasing on [−1,1][-1, 1] (Chapter 2, Fig 2.2 (ii)). Basics: trigonometric identities

  2. The inside is sec⁡2θ\sec 2\theta

    2x2−1=2cos⁡2θ−1=cos⁡2θ2x^2 - 1 = 2\cos^2\theta - 1 = \cos 2\theta⁠. Since 2θ∈(π2,π)2\theta \in \left(\frac{\pi}{2}, \pi\right)⁠, cos⁡2θ∈(−1,0)\cos 2\theta \in (-1, 0)⁠: it is not zero, so the fraction is defined, and

    12x2−1=1cos⁡2θ=sec⁡2θ<−1.\frac{1}{2x^2 - 1} = \frac{1}{\cos 2\theta} = \sec 2\theta < -1.
    sec⁡−1\sec^{-1} is defined only for inputs with absolute value at least 11⁠. The inside lies in (−∞,−1)(-\infty, -1)⁠, so yy is defined on the whole interval.
    cos⁡2θ=2cos⁡2θ−1\cos 2\theta = 2\cos^2\theta - 1 (Class XI); domain of sec⁡−1\sec^{-1} is R−(−1,1)\mathbf{R} - (-1, 1) (Chapter 2). Basics: trigonometric identities

  3. Simplify yy⁠, checking the range

    2θ∈(π2,π)2\theta \in \left(\frac{\pi}{2}, \pi\right) lies in [0,π]−{π2}[0, \pi] - \left\{\frac{\pi}{2}\right\}⁠, the principal range of sec⁡−1\sec^{-1}⁠. Therefore

    y=sec⁡−1(sec⁡2θ)=2θ=2cos⁡−1x.y = \sec^{-1}(\sec 2\theta) = 2\theta = 2\cos^{-1}x.
    sec⁡−1\sec^{-1} returns the unique angle in [0,π]−{π2}[0, \pi] - \left\{\frac{\pi}{2}\right\} with the given secant; 2θ2\theta is such an angle.
    sec⁡−1(sec⁡ϕ)=ϕ\sec^{-1}(\sec\phi) = \phi for ϕ∈[0,π]−{π2}\phi \in [0, \pi] - \left\{\frac{\pi}{2}\right\} (Chapter 2, Section 2.3). Basics: composite functions

  4. Differentiate

    dydx=2⋅−11−x2=−21−x2.\frac{dy}{dx} = 2\cdot\frac{-1}{\sqrt{1 - x^2}} = -\frac{2}{\sqrt{1 - x^2}}.
    The formula for ddxcos⁡−1x\frac{d}{dx}\cos^{-1}x needs −1<x<1-1 < x < 1⁠, which holds.
    Table after Example 24, Section 5.3.3. Lesson: derivatives of inverse trig functions

  5. Cross-check through cos⁡−1\cos^{-1} and the chain rule

    First, for ∣u∣≥1|u| \ge 1⁠, sec⁡−1u=cos⁡−11u\sec^{-1}u = \cos^{-1}\frac1u⁠: if ϕ=sec⁡−1u\phi = \sec^{-1}u⁠, then ϕ∈[0,π]\phi \in [0, \pi] and cos⁡ϕ=1u\cos\phi = \frac1u⁠, so ϕ\phi is the angle cos⁡−11u\cos^{-1}\frac1u⁠. Hence

    y=cos⁡−1(2x2−1).y = \cos^{-1}(2x^2 - 1).

    Now the chain rule, with inner function 2x2−12x^2 - 1 (derivative 4x4x⁠). Here 1−(2x2−1)2=4x2−4x41 - (2x^2 - 1)^2 = 4x^2 - 4x^4⁠, which is 4x2(1−x2)4x^2(1 - x^2)⁠, and its square root is 2x1−x22x\sqrt{1 - x^2} because x>0x > 0⁠. So

    dydx=−4x2x1−x2=−21−x2.\frac{dy}{dx} = \frac{-4x}{2x\sqrt{1 - x^2}} = -\frac{2}{\sqrt{1 - x^2}}.
    This route needs no angle substitution and agrees with Step 4. It uses x>0x > 0 to write 4x2=2x\sqrt{4x^2} = 2x⁠; for x<0x < 0 the sign would flip.
    Chain rule, Theorem 4; dducos⁡−1u=−11−u2\frac{d}{du}\cos^{-1}u = \frac{-1}{\sqrt{1 - u^2}}⁠, table after Example 24. Basics: derivative rules
0 π/4 π/2 π θ = cos⁻¹x lies in (π/4, π/2) 2θ lies in (π/2, π) range of sec⁻¹ [0, π] except π/2
For 0<x<120 < x < \frac{1}{\sqrt2}⁠, θ=cos⁡−1x\theta = \cos^{-1}x is on the gold arc (π4,π2)\left(\frac{\pi}{4}, \frac{\pi}{2}\right) and 2θ2\theta on the blue arc (π2,π)\left(\frac{\pi}{2}, \pi\right)⁠, inside the principal range of sec⁡−1\sec^{-1}⁠: [0,π][0, \pi] without π2\frac{\pi}{2} (violet; the open dot at the top is the excluded π2\frac{\pi}{2}⁠).
Result dydx=−21−x2,0<x<12.\frac{dy}{dx} = -\frac{2}{\sqrt{1 - x^2}}, \quad 0 < x < \frac{1}{\sqrt2}.

(Textbook answer: −21−x2-\dfrac{2}{\sqrt{1 - x^2}}⁠, the same.)

Check it with numbers

At x=0.5x = 0.5⁠: 2x2−1=−0.52x^2 - 1 = -0.5⁠, so the inside is −2-2 and y=sec⁡−1(−2)=2π3≈2.0943951y = \sec^{-1}(-2) = \frac{2\pi}{3} \approx 2.0943951⁠, which is 2cos⁡−10.52\cos^{-1}0.5⁠. With h=0.0001h = 0.0001⁠, y(0.5001)≈2.0941642y(0.5001) \approx 2.0941642 and y(0.4999)≈2.0946260y(0.4999) \approx 2.0946260⁠, and the difference quotient y(0.5001)−y(0.4999)0.0002\frac{y(0.5001) - y(0.4999)}{0.0002}⁠, computed from the unrounded values, is −2.3094011-2.3094011⁠; the formula gives −20.75≈−2.3094011-\frac{2}{\sqrt{0.75}} \approx -2.3094011⁠. ✓

Common mistakes
  1. Reversing the inequality wrongly. cos⁡−1\cos^{-1} is decreasing, so 0<x<120 < x < \frac{1}{\sqrt2} gives π4<θ<π2\frac{\pi}{4} < \theta < \frac{\pi}{2}⁠, not 0<θ<π40 < \theta < \frac{\pi}{4}⁠.
  2. Forgetting that π2\frac{\pi}{2} is excluded from the range of sec⁡−1\sec^{-1}⁠. Here 2θ2\theta never equals π2\frac{\pi}{2}⁠, because 2x2−1≠02x^2 - 1 \ne 0⁠.
  3. A sign error in the derivative of cos⁡−1\cos^{-1}⁠. It is −11−x2\frac{-1}{\sqrt{1 - x^2}}⁠; yy decreases as xx grows, so the answer is negative.
What to take away
  1. When an inverse function is not in the derivative table, simplify first, or rewrite it using sec⁡−1u=cos⁡−11u\sec^{-1}u = \cos^{-1}\frac1u⁠.
  2. 2x2−12x^2 - 1 with x=cos⁡θx = \cos\theta is cos⁡2θ\cos 2\theta⁠; keep track of whether the inverse function is increasing or decreasing when transferring an interval.

Every fact used, and where it comes from

FactUsed inSource
sec⁡−1:R−(−1,1)→[0,π]−{π2}\sec^{-1}: \mathbf{R} - (-1, 1) \to [0, \pi] - \left\{\frac{\pi}{2}\right\}⁠; cos⁡−1:[−1,1]→[0,π]\cos^{-1}: [-1, 1] \to [0, \pi]⁠, decreasingSteps 1 to 3, 5Chapter 2, Section 2.2 and Fig 2.2 (ii)
sec⁡−1(sec⁡ϕ)=ϕ\sec^{-1}(\sec\phi) = \phi on the principal rangeStep 3Chapter 2, Section 2.3
cos⁡2θ=2cos⁡2θ−1\cos 2\theta = 2\cos^2\theta - 1⁠; sec⁡ϕ=1cos⁡ϕ\sec\phi = \frac{1}{\cos\phi}Step 2Class XI, Trigonometric Functions
ddxcos⁡−1x=−11−x2\frac{d}{dx}\cos^{-1}x = \frac{-1}{\sqrt{1 - x^2}}⁠, −1<x<1-1 < x < 1Steps 4, 5Table after Example 24, Section 5.3.3 (Lesson: derivatives of inverse trig functions)
Chain ruleStep 5Theorem 4, Section 5.3.1
Numerical valuesCheckRecomputed in the site's verification script
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