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Chapter 5 · Continuity and Differentiability

Exercise 5.7 · Q8: Second derivative of tan⁡−1x\tan^{-1}x

Exercise 5.7Second order derivative · Section 5.7
The question

Find the second order derivatives of the functions given in Exercises 1 to 10.

8. tan⁡−1x\tan^{-1}x

Exam answer

Let y=tan⁡−1xy = \tan^{-1}x⁠. Then

dydx=11+x2=(1+x2)−1\frac{dy}{dx} = \frac{1}{1 + x^2} = (1 + x^2)^{-1}

By the chain rule,

d2ydx2=−(1+x2)−2⋅2x\frac{d^2y}{dx^2} = -(1 + x^2)^{-2}\cdot 2x

∴ d2ydx2=−2x(1+x2)2\therefore\ \dfrac{d^2y}{dx^2} = -\dfrac{2x}{(1 + x^2)^2}

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

The one idea this problem is built on

The first derivative of tan⁡−1x\tan^{-1}x is the standard result 11+x2\frac{1}{1+x^2}⁠. To differentiate it again, write it as a power of an inner function,

11+x2=(1+x2)−1,\frac{1}{1+x^2} = (1+x^2)^{-1},

and use the chain rule: outer function "power −1-1", inner function 1+x21 + x^2⁠.

Given, to find, and the plan
Given
y=tan⁡−1xy = \tan^{-1}x⁠, defined and differentiable for every real xx⁠, with values in (−π2,π2)\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)⁠.
To find
d2ydx2\dfrac{d^2y}{dx^2}⁠.
Plan
(1) State the domain. (2) dydx=11+x2\frac{dy}{dx} = \frac{1}{1+x^2} from Section 5.3.3. (3) Chain rule on (1+x2)−1(1+x^2)^{-1}⁠. (4) Read the sign of the answer on the graph.

Basics you need

1Derivative of tan⁡−1x\tan^{-1}x

ddxtan⁡−1x=11+x2\frac{d}{dx}\tan^{-1}x = \frac{1}{1+x^2} for every real xx (Section 5.3.3, table of derivatives of inverse trigonometric functions; domain R\mathbf{R}⁠). Lesson: derivatives of inverse trig functions

2Chain rule (Theorem 4)

For y=f(g(x))y = f(g(x))⁠: dydx=f′(g(x))⋅g′(x)\frac{dy}{dx} = f'(g(x))\cdot g'(x)⁠. With f(u)=u−1f(u) = u^{-1} this gives

ddx(g(x))−1=−(g(x))−2 g′(x).\frac{d}{dx}\big(g(x)\big)^{-1} = -\big(g(x)\big)^{-2}\,g'(x).

Lesson: chain rule

3Power rule

dduun=nun−1\frac{d}{du}u^n = nu^{n-1} (Table 5.3), here with n=−1n = -1 and n=2n = 2⁠. Basics: derivative rules

4Second order derivative (Section 5.7)

d2ydx2=ddx(dydx)\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right)⁠. Lesson: second order derivative

Solution

  1. Name the function and its domain

    Let y=tan⁡−1xy = \tan^{-1}x⁠. Its domain is all of R\mathbf{R} (Chapter 2: tan⁡−1:R→(−π2,π2)\tan^{-1} : \mathbf{R} \to \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)⁠). Also 1+x2≥1>01 + x^2 \ge 1 \gt 0 for every real xx⁠, so the denominators below are never zero. The answer holds for every x∈Rx \in \mathbf{R}⁠.

    Before dividing by 1+x21 + x^2 or (1+x2)2(1 + x^2)^2 we make sure they cannot be zero. Since x2≥0x^2 \ge 0⁠, adding 1 keeps them at least 1.
    Principal value branch of tan⁡−1\tan^{-1} (Chapter 2, Section 2.2). Lesson: derivatives of inverse trig functions

  2. First derivative: the standard result

    dydx=ddx(tan⁡−1x)=11+x2.\frac{dy}{dx} = \frac{d}{dx}\left(\tan^{-1}x\right) = \frac{1}{1+x^2}.
    The book lists this in the table at the end of Section 5.3.3, with domain R\mathbf{R}⁠. It comes out exactly as Example 24 does for sin⁡−1x\sin^{-1}x⁠: from x=tan⁡yx = \tan y⁠, differentiating gives 1=sec⁡2y dydx1 = \sec^2 y\,\frac{dy}{dx}⁠, so dydx=1sec⁡2y\frac{dy}{dx} = \frac{1}{\sec^2 y}⁠, and sec⁡2y=1+tan⁡2y=1+x2\sec^2 y = 1 + \tan^2 y = 1 + x^2⁠.
    ddxtan⁡−1x=11+x2\frac{d}{dx}\tan^{-1}x = \frac{1}{1+x^2} (Section 5.3.3).

  3. Second derivative: chain rule on (1+x2)−1(1+x^2)^{-1}

    Write dydx=(1+x2)−1\frac{dy}{dx} = (1+x^2)^{-1}⁠. Inner function u=1+x2u = 1 + x^2⁠, with dudx=2x\frac{du}{dx} = 2x (power rule; the constant 1 gives 0). Outer function u−1u^{-1}⁠, with derivative −u−2-u^{-2} (power rule, n=−1n = -1⁠). By the chain rule,

    d2ydx2=−(1+x2)−2⋅2x=−2x(1+x2)2.\begin{aligned} \frac{d^2y}{dx^2} &= -(1+x^2)^{-2}\cdot 2x \\ &= -\frac{2x}{(1+x^2)^2}. \end{aligned}

    So d2ydx2=−2x(1+x2)2\dfrac{d^2y}{dx^2} = -\dfrac{2x}{(1+x^2)^2} for every real xx⁠.

    11+x2\frac{1}{1+x^2} is "one over something", that is, the something to the power −1-1⁠. The power rule handles the outside and the chain rule multiplies by the derivative of the inside, 2x2x⁠. The quotient rule gives the same: 0⋅(1+x2)−1⋅2x(1+x2)2\frac{0\cdot(1+x^2) - 1\cdot 2x}{(1+x^2)^2}⁠.
    Chain rule (Theorem 4, Section 5.3.1); power rule (Table 5.3). Lesson: chain rule

  4. Read the sign: where the graph bends up and down

    The denominator (1+x2)2(1+x^2)^2 is always positive, so the sign of d2ydx2\frac{d^2y}{dx^2} is the sign of −2x-2x⁠:

    • for x<0x \lt 0⁠, y′′>0y'' \gt 0⁠: the slope 11+x2\frac{1}{1+x^2} is increasing, and the graph bends upward;
    • for x>0x \gt 0⁠, y′′<0y'' \lt 0⁠: the slope is decreasing, and the graph bends downward;
    • at x=0x = 0⁠, y′′=0y'' = 0 and the bending changes direction; there the slope is largest, equal to 1.
    x y −4 −2 2 4 y = π/2 y = −π/2 bends upward y'' > 0 for x < 0 bends downward y'' < 0 for x > 0 (0, 0)
    y=tan⁡−1xy = \tan^{-1}x for −4.5≤x≤4.5-4.5 \le x \le 4.5 (one unit = 55 px on both axes), between the dashed lines y=±π2y = \pm\frac{\pi}{2} that it never reaches. Green part: y′′>0y'' \gt 0 (bends upward). Red part: y′′<0y'' \lt 0 (bends downward). The bend changes at the origin, where y′′=0y'' = 0⁠.
    The sign of the second derivative tells you which way a graph bends; Chapter 6 uses this to classify turning points. Seeing it here also confirms the minus sign in the answer: the curve flattens out as xx grows.
Result

For y=tan⁡−1xy = \tan^{-1}x (all real xx⁠):

dydx=11+x2,d2ydx2=−2x(1+x2)2.\begin{aligned} \frac{dy}{dx} &= \frac{1}{1+x^2}, \\ \frac{d^2y}{dx^2} &= -\frac{2x}{(1+x^2)^2}. \end{aligned}

(Textbook answer: −2x(1+x2)2-\dfrac{2x}{(1+x^2)^2}⁠.)

Check it with numbers

At x=1x = 1⁠: formula y′′=−2(1+1)2=−24=−0.5y'' = -\frac{2}{(1+1)^2} = -\frac{2}{4} = -0.5⁠. With h=0.001h = 0.001⁠, tan⁡−11.001≈0.7858979\tan^{-1}1.001 \approx 0.7858979⁠, tan⁡−11=π4≈0.7853982\tan^{-1}1 = \frac{\pi}{4} \approx 0.7853982⁠, tan⁡−10.999≈0.7848979\tan^{-1}0.999 \approx 0.7848979⁠, and

tan⁡−11.001−2⋅π4+tan⁡−10.9990.000001≈−0.5000000.\begin{aligned} &\frac{\tan^{-1}1.001 - 2\cdot\frac{\pi}{4} + \tan^{-1}0.999}{0.000001} \\ &\approx -0.5000000. \end{aligned}

The first difference tan⁡−11.001−tan⁡−10.9990.002\frac{\tan^{-1}1.001 - \tan^{-1}0.999}{0.002} is about 0.50000010.5000001⁠, against 11+1=0.5\frac{1}{1+1} = 0.5⁠. ✓

Symmetry check: tan⁡−1(−x)=−tan⁡−1x\tan^{-1}(-x) = -\tan^{-1}x is an odd function, so its second derivative must be odd too; −2x(1+x2)2-\frac{2x}{(1+x^2)^2} is indeed odd.

Common mistakes
  1. Forgetting the inner derivative 2x2x⁠: −1(1+x2)2-\frac{1}{(1+x^2)^2} is wrong.
  2. Writing tan⁡−1x\tan^{-1}x as 1tan⁡x\frac{1}{\tan x}⁠. The −1-1 means inverse function, not reciprocal; 1tan⁡x=cot⁡x\frac{1}{\tan x} = \cot x is a different function.
  3. Losing the minus sign that the power −1-1 brings down.
What to take away
  1. To differentiate 1g(x)\frac{1}{g(x)}⁠, write (g(x))−1(g(x))^{-1} and use the chain rule: −g′(x)(g(x))2-\frac{g'(x)}{(g(x))^2}⁠.
  2. tan⁡−1x\tan^{-1}x bends upward for x<0x \lt 0 and downward for x>0x \gt 0⁠, changing at the origin.

Every fact used, and where it comes from

FactUsed inSource
tan⁡−1:R→(−π2,π2)\tan^{-1} : \mathbf{R} \to \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)Step 1Chapter 2, Section 2.2 (principal value branches)
ddxtan⁡−1x=11+x2\frac{d}{dx}\tan^{-1}x = \frac{1}{1+x^2}⁠, x∈Rx \in \mathbf{R}Step 2Section 5.3.3 (Lesson: derivatives of inverse trig functions)
Chain ruleStep 3Theorem 4, Section 5.3.1 (Lesson: chain rule)
dduun=nun−1\frac{d}{du}u^n = nu^{n-1}Step 3Table 5.3 (Basics: derivative rules)
d2ydx2=ddx(dydx)\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right)Step 3Section 5.7 (Lesson: second order derivative)
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