Implicit differentiation and inverse trigonometric functions, question 23

Topic: Implicit differentiation and inverse trigonometric functions
Difficulty: Challenging
Suggested time: 5 min
Marks: 3
Theory: UBC Math 100 §4.7 CLP-1 §2.12Key results

Let $y = \arccos x$ for $-1 < x < 1$ . Starting from $\cos y = x$ , use implicit differentiation to show that

\frac{d}{dx}\arccos x = -\frac{1}{\sqrt{1 - x^{2}}}

Explain how the range of arccos fixes the sign of the square root, then compare against the model solution.

Your answer

Theory UBC Math 100 §4.7 CLP-1 §2.12Key results

Worked solution

Strategy

Turn the inverse function into an equation about cos, differentiate it implicitly, and then rewrite sin y in terms of x. The range of arccos decides the sign of the root.

Working

$y = \arccos x \quad\Longleftrightarrow\quad \cos y = x, \quad 0 \le y \le \pi$
The range of arccos fixes which angle y is.
$-\sin y\,\frac{dy}{dx} = 1 \quad\Longrightarrow\quad \frac{dy}{dx} = -\frac{1}{\sin y}$
The derivative of cos is −sin, and the chain rule attaches dy/dx.
$\sin y = \sqrt{1 - \cos^{2}y} = \sqrt{1 - x^{2}}$
On [0, π] the sine is never negative, which picks the positive root.
$\frac{dy}{dx} = -\frac{1}{\sqrt{1 - x^{2}}}, \qquad -1 < x < 1$
At x = ±1, sin y = 0 and the formula breaks down: the graph is vertical at its endpoints.

Answer

\frac{d}{dx}\arccos x = -\frac{1}{\sqrt{1 - x^{2}}}

What earns the marks

Rewrites y = arccos x as cos y = x with 0 ≤ y ≤ π, and differentiates both sides to get −sin y · dy/dx = 1.
Writes sin y as √(1 − x²), using sin²y + cos²y = 1 and the fact that sin y ≥ 0 for y in [0, π].
Concludes dy/dx = −1/√(1 − x²), noting that x = ±1 is excluded because sin y = 0 there.

Check

Since arcsin x + arccos x = π/2 on [−1, 1], their derivatives must be negatives of each other, and −1/√(1 − x²) is minus the derivative of arcsin.

Where this goes wrong

Differentiating cos y as sin y · dy/dx, which loses the minus sign and produces the derivative of arcsin. The minus sign is what makes arccos a decreasing function.

Leans on

The derivative of cosine, from Exponential and trigonometric functions

Try a similar question

Let

f(x) = \arccos\!\left(\dfrac{x}{2}\right)

. Find

f'(1)

My working

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Implicit differentiation and inverse trigonometric functions practice question 23