Implicit differentiation and inverse trigonometric functions, question 29

Topic: Implicit differentiation and inverse trigonometric functions
Difficulty: Capstone
Suggested time: 9 min
Marks: 5
Theory: UBC Math 100 §4.5 CLP-1 §2.11Key results

The equation below defines $y$ as a differentiable function of $x$ for every real $x$ . Find $\dfrac{d^{2}y}{dx^{2}}$ at $x = 1 + \dfrac{\pi}{4}$ .

y + \arctan y = x

Your answer

An exact value, or a decimal to three places.

How sure are you?

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Theory UBC Math 100 §4.5 CLP-1 §2.11Key results

Worked solution

Strategy

First find the point: arctan 1 = π/4 gives y = 1, and it is the only y because y + arctan y is increasing. Implicit differentiation then gives y′ as a formula in y, and differentiating that formula again needs the chain rule once more, because y still depends on x; the numbers go in only at the end.

Working

$x = 1 + \tfrac{\pi}{4}:\qquad y + \arctan y = 1 + \tfrac{\pi}{4} \quad\Longrightarrow\quad y = 1$
arctan 1 = π/4, so y = 1 works; it is the only solution, because y + arctan y is increasing.
$y' + \frac{y'}{1 + y^{2}} = 1$
Differentiate both sides; the chain rule gives (arctan y)′ = y′/(1 + y²).
$y' = \frac{1}{1 + \frac{1}{1 + y^{2}}} = \frac{1 + y^{2}}{2 + y^{2}}$
Factor out y′. The bracket 1 + 1/(1 + y²) is never 0, so y′ exists at every point.
$y = 1:\qquad y' = \frac{1 + 1}{2 + 1} = \frac{2}{3}$
The slope at the point.
$y'' = \frac{2y\,y'\,(2 + y^{2}) - (1 + y^{2})\cdot 2y\,y'}{(2 + y^{2})^{2}} = \frac{2y\,y'}{(2 + y^{2})^{2}}$
The quotient rule, with (y²)′ = 2y y′; the y³ terms cancel.
$y'' = \frac{2\cdot 1\cdot \frac{2}{3}}{3^{2}} = \frac{4}{27}$
Now substitute y = 1 and y′ = 2/3.

Answer

\left.\frac{d^{2}y}{dx^{2}}\right|_{x = 1 + \pi/4} = \frac{4}{27}

Check

Solving y + arctan y = x numerically, y′ = (1 + y²)/(2 + y²) rises from about 0.6519 at x = 1.6854 to 0.6815 at x = 1.8854, a rate of about 0.148.

Where this goes wrong

Treating y as a constant when differentiating y′ = (1 + y²)/(2 + y²) a second time, which gives 0. y changes with x, so the derivative of y² is 2y y′; substitute y = 1 and y′ = 2/3 only after differentiating, and remember that 1 + π/4 is the value of x, not of y.