Derivatives of Inverse Trigonometric Functions Class 12: Formulas & Solved Problems

Inverse trigonometric functions — sin⁻¹x, cos⁻¹x, tan⁻¹x and their counterparts — look intimidating at first, but their derivatives follow a small, memorable set of formulas. This topic is a guaranteed scorer in CBSE board exams and CUET Maths papers, especially when combined with clever trigonometric substitutions. Here’s everything you need, from derivation to solved problems.

The Standard Derivative Formulas

Before diving into derivations, memorize this table — it’s the foundation for every question in this topic:

Notice the pattern: sin⁻¹x and cos⁻¹x have the same magnitude of derivative but opposite signs, while tan⁻¹x has a completely different structure with no square root and a domain of all real numbers.

Deriving the Formula for sin⁻¹x

Step-by-Step Derivation

Let y = sin⁻¹x, so by definition, x = sin y.

Differentiating both sides with respect to x:

1 = cos y × (dy/dx)

dy/dx = 1 / cos y

Now, since y ∈ (−π/2, π/2), cos y is always positive, so we can write cos y = √(1 − sin²y) = √(1 − x²) (using sin y = x).

dy/dx = 1 / √(1 − x²), valid for x ∈ (−1, 1)

This derivation pattern — differentiate implicitly, then rewrite in terms of x using the appropriate identity — is exactly how all inverse trig derivatives are derived, so understanding this one example unlocks the rest.

Solved Example 1: Basic Application

Differentiate f(x) = sin⁻¹x + sin⁻¹(√(1−x²)), for 0 < x < 1.

Using the standard formulas: d/dx[sin⁻¹x] = 1/√(1−x²)

For the second term, let g(x) = √(1−x²). By the chain rule, d/dx[sin⁻¹(g(x))] = [1/√(1−g²)] × g′(x). Since g² = 1−x², this simplifies to 1/√(x²) = 1/x (for the relevant domain), and g′(x) = −x/√(1−x²).

Combining terms and simplifying (a standard CBSE technique) shows this expression is actually constant over the given domain — differentiating it gives 0. This is a classic trick question: always simplify before you differentiate blindly!

Solved Example 2: Using Trigonometric Substitution

Find dy/dx if y = tan⁻¹((3x − x³)/(1 − 3x²)), for −1/√3 < x < 1/√3.

This expression matches the triple-angle identity for tangent: tan(3θ) = (3tanθ − tan³θ)/(1 − 3tan²θ).

Substitute x = tan θ, so θ = tan⁻¹x. Then:

y = tan⁻¹(tan 3θ) = 3θ = 3 tan⁻¹x

dy/dx = 3 / (1 + x²)

This substitution technique — recognizing double-angle or triple-angle identities hidden inside inverse trig expressions — is one of the most tested skills in this section. Common substitutions to remember: x = sin θ or x = cos θ for expressions with √(1−x²), x = tan θ for expressions with (1+x²), and x = sec θ for expressions with √(x²−1).

Solved Example 3: A Compound Expression

Find dy/dx if y = sin⁻¹(2x/(1+x²)), for −1 < x < 1.

Substitute x = tan θ:

2x/(1+x²) = 2tanθ/(1+tan²θ) = sin(2θ) [using the double angle identity]

So y = sin⁻¹(sin 2θ) = 2θ = 2 tan⁻¹x, because for −1 < x < 1 we have 2θ ∈ (−π/2, π/2), which is inside the principal range of sin⁻¹.

dy/dx = 2 / (1 + x²), for −1 < x < 1

Why the domain matters: for x > 1, 2θ lies in (π/2, π), so sin⁻¹(sin 2θ) = π − 2θ and the derivative becomes −2 / (1 + x²). For x < −1 it is also −2 / (1 + x²), and the function is not differentiable at x = ±1.

Notice how the same substitution trick (x = tan θ) simplifies a scary-looking inverse trig expression into something you can differentiate in one line.

Why These Substitutions Work

The substitutions x = sinθ, x = tanθ, or x = secθ are chosen specifically to match a known trigonometric identity buried in the expression — usually a double-angle or triple-angle formula. Once substituted, the inverse trig function often collapses to a simple multiple of θ (like 2θ or 3θ), and since θ = (the corresponding inverse trig function of x), differentiating becomes trivial. Recognizing which substitution to use comes with practice — look for structures like 2x/(1+x²), (1−x²)/(1+x²), or (3x−x³)/(1−3x²) as strong hints.

A Note on Domain Restrictions

CBSE examiners frequently test whether you remember that these derivative formulas only hold within specific domains. sin⁻¹x and cos⁻¹x are defined for x ∈ [−1, 1], but their derivatives are only valid on the open interval (−1, 1) — at the endpoints x = 1 and x = −1, the derivative formula involves division by zero (since √(1−x²) = 0 there), so the function is not differentiable at those boundary points, even though it is continuous. tan⁻¹x, by contrast, is defined and differentiable across all real numbers, since 1 + x² can never be zero. Always state the domain alongside your final derivative in board answers — examiners award marks for this precision.

Tips for Scoring Full Marks in This Topic

1. Memorize the three core formulas — everything else builds on sin⁻¹x, cos⁻¹x, and tan⁻¹x.

2. Always check the domain before differentiating — the formulas only hold within specific ranges.

3. Look for hidden identities before applying the chain rule directly — many “hard” problems become one-line answers after the right substitution.

4. Simplify algebraically first whenever you see combinations like sin⁻¹x + sin⁻¹(√(1−x²)), which equals π/2 for 0 ≤ x ≤ 1 — they’re often disguised constants.

5. Practice both derivation and application — CBSE tests both “find the derivative of sin⁻¹x” (recall) and complex substitution problems (application) in the same paper.

Master Inverse Trig Derivatives With Targeted Practice

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Test Your Knowledge

Q1.The derivative of sin⁻¹(x) with respect to x for |x| < 1 is:

−1 / √(1 − x²)1 / √(1 − x²)1 / (1 + x²)1 / (x √(x² − 1))

Q2.The derivative of tan⁻¹(x) with respect to x for all real x is:

−1 / (1 + x²)1 / √(1 + x²)1 / (1 + x²)sec²(x)

Q3.How is the derivative of cos⁻¹(x) related to the derivative of sin⁻¹(x)?

It is the reciprocal: 1 / (d/dx [sin⁻¹(x)])They are identicalTheir product equals 1It is the exact negative: d/dx [cos⁻¹(x)] = −d/dx [sin⁻¹(x)]