Differentiating Inverse Trigonometric Functions: Study Guide (Calculus AB 3.4)

Course scope. This topic is shared by Calculus AB and Calculus BC. Everything here is examinable for both.

A note on notation

• arcsin x and sin⁻¹ x mean the same thing: the angle, in radians, whose sine is x. The same goes for arccos x (cos⁻¹ x) and arctan x (tan⁻¹ x).
• sin⁻¹ x is not 1/sin x. The reciprocal of sin x is csc x. A power such as sin²x is a different notation from sin⁻¹ x.
• All angles are in radians. The derivative formulas below are only true in radians.

This page writes “arcsin” rather than “sin⁻¹” to avoid mixing up inverse and reciprocal.

Why the ranges are restricted

Sine is not one-to-one: sin(π/6) = sin(5π/6) = 1/2. To have an inverse, sine must be restricted to an interval where it is one-to-one. The standard choices are:

Function	Domain (inputs x)	Range (output angles)	The original function is restricted to
arcsin x	−1 ≤ x ≤ 1	−π/2 ≤ y ≤ π/2	sin on [−π/2, π/2], where it increases
arccos x	−1 ≤ x ≤ 1	0 ≤ y ≤ π	cos on [0, π], where it decreases
arctan x	all real x	−π/2 < y < π/2	tan on (−π/2, π/2), where it increases

The ranges matter in the derivations below. On [−π/2, π/2], cos y is never negative. On [0, π], sin y is never negative. Those facts fix the sign of each square root.

Deriving the derivative of arcsin x

Let y = arcsin x, for −1 < x < 1. By the definition of the inverse,

sin y = x, with −π/2 < y < π/2.

Differentiate both sides with respect to x, using implicit differentiation (Topic 3.2):

cos y · dy/dx = 1, so dy/dx = 1/cos y.

Now write cos y in terms of x. From sin²y + cos²y = 1, cos y = ±√(1 − sin²y) = ±√(1 − x²). Because y lies between −π/2 and π/2, cos y > 0, so take the positive root:

d/dx arcsin x = 1/√(1 − x²), for −1 < x < 1.

A right triangle shows the same step without the identity (Figure 1). If sin y = x, draw a triangle with hypotenuse 1 and opposite side x. Pythagoras gives the adjacent side √(1 − x²), so cos y = √(1 − x²). (The triangle shows the case 0 < x < 1; the identity argument covers all of −1 < x < 1.)

The same result from Topic 3.3. Sine (restricted) and arcsin are inverses, so (arcsin)′(x) = 1/sin′(arcsin x) = 1/cos(arcsin x) = 1/√(1 − x²). This is the inverse-function rule g′(x) = 1/f′(g(x)) with f = sin.

Deriving the derivative of arctan x

Let y = arctan x. Then tan y = x, with −π/2 < y < π/2. Differentiate implicitly:

sec²y · dy/dx = 1, so dy/dx = 1/sec²y.

Use the identity sec²y = 1 + tan²y = 1 + x² (or the right-hand triangle in Figure 1):

d/dx arctan x = 1/(1 + x²), for every real x.

The denominator 1 + x² is never 0, so arctan is differentiable everywhere. Its slope is largest at x = 0, where it equals 1, and it gets close to 0 as x grows large in either direction. That matches the graph of arctan, which levels off towards the horizontal asymptotes y = ±π/2.

The other inverse trigonometric functions

arccos. For every x in [−1, 1], arcsin x + arccos x = π/2. (The two angles are complementary.) Differentiating, d/dx arccos x = 0 − 1/√(1 − x²):

d/dx arccos x = −1/√(1 − x²), for −1 < x < 1.

The negative sign makes sense: arccos is a decreasing function.

The remaining three follow from the same method. They are less common, and their formulas depend on the range convention (the ones below use the usual ranges: arccot from 0 to π; arcsec and arccsc built from arccos(1/x) and arcsin(1/x)).

Function	Derivative	Valid for
arcsin x	1/√(1 − x²)	−1 < x < 1
arccos x	−1/√(1 − x²)	−1 < x < 1
arctan x	1/(1 + x²)	all real x
arccot x	−1/(1 + x²)	all real x
arcsec x	1/(|x|√(x² − 1))	|x| > 1
arccsc x	−1/(|x|√(x² − 1))	|x| > 1

Where the derivatives fail. At x = ±1 the denominator √(1 − x²) is 0. The graphs of arcsin and arccos have vertical tangents at their endpoints, because sine and cosine have horizontal tangents at the matching angles (Topic 3.3). So arcsin and arccos are differentiable only on the open interval −1 < x < 1.

Using the chain rule

Most exam questions put something inside the inverse function. Combine each formula with the chain rule (Topic 3.1). If u is a differentiable function of x:

• d/dx arcsin u = u′/√(1 − u²)
• d/dx arccos u = −u′/√(1 − u²)
• d/dx arctan u = u′/(1 + u²)

Two checks: the whole inner function is squared (not just x), and the derivative of the inner function goes on top.

Worked example 1: arctan with a chain

Question. Let h(x) = arctan(x² − 1). Find h′(x). Then find the equation of the tangent line to the graph of h at x = 1.

1. Identify the inner function. u = x² − 1, so u′ = 2x.
2. Apply the chain-rule form. h′(x) = u′/(1 + u²) = 2x/(1 + (x² − 1)²). Expanded, the denominator is x⁴ − 2x² + 2. Either form is fine.
3. Point on the graph. h(1) = arctan(0) = 0. The point is (1, 0).
4. Slope. h′(1) = 2(1)/(1 + 0²) = 2.
5. Tangent line. y = 2(x − 1).

Check. At x = 1.01, h(1.01) = arctan(0.0201) ≈ 0.0201, and the tangent line gives 2(0.01) = 0.02. These agree closely.

One more value. h′(0) = 0, because the factor 2x is 0 there. The graph of h has a horizontal tangent at (0, −π/4), the lowest point of the graph, since x² − 1 is smallest at x = 0.

Worked example 2: arcsin inside a product

Question. Let y = x · arcsin(2x). Find dy/dx and evaluate it at x = 1/4.

1. Where is y differentiable? arcsin(2x) needs −1 ≤ 2x ≤ 1, and its derivative needs −1 < 2x < 1. So work on −1/2 < x < 1/2. The value x = 1/4 is inside.

2. Product rule. dy/dx = (1) · arcsin(2x) + x · d/dx arcsin(2x).

3. Chain rule on arcsin(2x). u = 2x, u′ = 2, so d/dx arcsin(2x) = 2/√(1 − 4x²).

4. Combine. dy/dx = arcsin(2x) + 2x/√(1 − 4x²).

5. Evaluate at x = 1/4. arcsin(1/2) = π/6. And 1 − 4(1/16) = 3/4, so √(3/4) = √3/2. The second term is (2/4)/(√3/2) = (1/2)(2/√3) = 1/√3 = √3/3.

   dy/dx at x = 1/4 equals π/6 + √3/3 (about 1.101).

Check. A symmetric difference using y at x = 0.251 and 0.249 gives about 1.101, which agrees.

Watch the square. The inner function is 2x, so u² = 4x², not 2x². Squaring only the x is one of the most common slips in this topic.

Worked example 3: using the inverse-function rule at a point

Question. Without using the arccos formula, find the derivative of arccos x at x = 1/2. Then confirm with the formula.

1. Matching input. arccos(1/2) = π/3, because cos(π/3) = 1/2 and π/3 is in [0, π].
2. Derivative of the original function there. The original function is cos, restricted to [0, π]. Its derivative is −sin, and −sin(π/3) = −√3/2.
3. Reciprocal (Topic 3.3). The derivative of arccos at 1/2 is 1/(−√3/2) = −2/√3, which is −2√3/3.

Confirm. −1/√(1 − 1/4) = −1/(√3/2) = −2/√3. The two methods agree, and the answer is negative, as it must be for a decreasing function.

Common misconceptions

• Reading sin⁻¹ x as 1/sin x. That is csc x, whose derivative is −csc x cot x. arcsin is a different function.
• Mixing up the formulas. arcsin and arccos have a square root; arctan does not. Ask: “Is there a square root in this one?” Of these three, only the sine and cosine versions have one.
• Forgetting the minus sign for arccos. arccos decreases, so its derivative is negative.
• Dropping the chain rule. d/dx arctan(5x) is 5/(1 + 25x²), not 1/(1 + 25x²).
• Squaring only part of the inner function. For arcsin(2x), the denominator is √(1 − 4x²), not √(1 − 2x²).
• Using degrees. arcsin(1/2) is π/6, not 30, in every calculus formula.
• Evaluating outside the domain. arcsin(1.2) is undefined, because no angle has sine 1.2. And arcsin is not differentiable at x = ±1.
• Picking the wrong sign of the root. cos(arcsin x) = +√(1 − x²), because arcsin gives angles where cosine is not negative.

Where this leads

Topic 3.5 asks you to choose the right procedure when a function mixes these derivatives with products, quotients and compositions. Continue with Selecting Procedures for Calculating Derivatives. In Unit 6 you will read these formulas backwards: 1/(1 + x²) and 1/√(1 − x²) are the derivatives of arctan x and arcsin x, so they have those functions as antiderivatives. Students in either course can return to the Calculus AB roadmap or the Calculus BC roadmap.

Try the practice questions now, then use the revision notes and the checklist to consolidate.