Derivatives of Inverse Trigonometric Functions

4 min read

Reference: Stewart §6.6

Why Do These Derivatives Contain Square Roots?

The derivatives of $\sin^{-1}x$ and $\cos^{-1}x$ have $\sqrt{1-x^2}$ in them. This is not coincidental: it comes directly from the Pythagorean identity $\sin^2\theta + \cos^2\theta = 1$. When we solve for $\cos\theta$ in terms of $\sin\theta = x$, we get $\cos\theta = \sqrt{1-x^2}$.

These derivatives are essential for integration. Whenever you see $\frac{1}{\sqrt{1-x^2}}$ or $\frac{1}{1+x^2}$ in an integral, you’re looking at an inverse trig antiderivative.

📋 Before You Start: Check Your Prerequisites

Implicit differentiation: If $x^2 + y^2 = 1$, find $\frac{dy}{dx}$ implicitly.
- If unsure: Review Implicit Differentiation
Pythagorean identity: What is $\sin^2\theta + \cos^2\theta$?
- Answer: 1
Chain rule: If $y = \sin(3x)$, what is $\frac{dy}{dx}$?
- Answer: $3\cos(3x)$

Prerequisite Map

Quick Reference

Property	Value
Concept	Inverse Function Derivatives
Chapter	3.6
Difficulty	Intermediate
Time	~25 minutes

Key Concepts

The Three Essential Formulas

$$\boxed{\frac{d}{dx}(\sin^{-1}x) = \frac{1}{\sqrt{1-x^2}}, \quad -1 < x < 1}$$

$$\boxed{\frac{d}{dx}(\cos^{-1}x) = -\frac{1}{\sqrt{1-x^2}}, \quad -1 < x < 1}$$

$$\boxed{\frac{d}{dx}(\tan^{-1}x) = \frac{1}{1+x^2}, \quad x \in \mathbb{R}}$$

Deriving the arcsin Derivative

Let $y = \sin^{-1}x$. Then: $$\sin y = x \quad \text{and} \quad -\frac{\pi}{2} \leq y \leq \frac{\pi}{2}$$

Differentiating implicitly: $$\cos y \cdot \frac{dy}{dx} = 1$$

$$\frac{dy}{dx} = \frac{1}{\cos y}$$

Since $-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}$, we have $\cos y \geq 0$. Using $\sin^2 y + \cos^2 y = 1$:

$$\cos y = \sqrt{1 - \sin^2 y} = \sqrt{1 - x^2}$$

Therefore: $$\frac{d}{dx}(\sin^{-1}x) = \frac{1}{\sqrt{1-x^2}}$$

Deriving the arctan Derivative

Let $y = \tan^{-1}x$. Then: $$\tan y = x \quad \text{and} \quad -\frac{\pi}{2} < y < \frac{\pi}{2}$$

Differentiating implicitly: $$\sec^2 y \cdot \frac{dy}{dx} = 1$$

$$\frac{dy}{dx} = \frac{1}{\sec^2 y} = \cos^2 y$$

Using $\sec^2 y = 1 + \tan^2 y$: $$\frac{dy}{dx} = \frac{1}{1 + \tan^2 y} = \frac{1}{1 + x^2}$$

The Chain Rule Forms

For a differentiable function $u = g(x)$:

Formula	Chain Rule Version
$\frac{d}{dx}(\sin^{-1}u)$	$\frac{1}{\sqrt{1-u^2}} \cdot \frac{du}{dx}$
$\frac{d}{dx}(\cos^{-1}u)$	$-\frac{1}{\sqrt{1-u^2}} \cdot \frac{du}{dx}$
$\frac{d}{dx}(\tan^{-1}u)$	$\frac{1}{1+u^2} \cdot \frac{du}{dx}$

Complete Table (Including sec, csc, cot)

Function	Derivative	Domain of $x$
$\sin^{-1}x$	$\frac{1}{\sqrt{1-x^2}}$	$(-1, 1)$
$\cos^{-1}x$	$-\frac{1}{\sqrt{1-x^2}}$	$(-1, 1)$
$\tan^{-1}x$	$\frac{1}{1+x^2}$	$\mathbb{R}$
$\cot^{-1}x$	$-\frac{1}{1+x^2}$	$\mathbb{R}$
$\sec^{-1}x$	$\frac{1}{\vert x\vert \sqrt{x^2-1}}$	$\vert x\vert > 1$
$\csc^{-1}x$	$-\frac{1}{\vert x\vert \sqrt{x^2-1}}$	$\vert x\vert > 1$

Notice the patterns:

$\sin^{-1}$ and $\cos^{-1}$ have opposite signs (sum to $\pi/2$)
$\tan^{-1}$ and $\cot^{-1}$ have opposite signs (sum to $\pi/2$)
$\sec^{-1}$ and $\csc^{-1}$ have opposite signs

Visualizing the Domains

sin⁻¹x domain:        ←───|═══════|───→
                         -1       1

tan⁻¹x domain:        ←═══════════════→
                       all real numbers

sec⁻¹x domain:        ←═══|       |═══→
                         -1       1
                      (excludes middle)

Practice Problems

Level 1 Direct Formula Application

Find $\frac{d}{dx}[\tan^{-1}(3x)]$.

Thought Process

Use the chain rule with $u = 3x$.

The derivative of $\tan^{-1}u$ is $\frac{1}{1+u^2}$, and multiply by $\frac{du}{dx} = 3$.

Show Answer

$$\frac{d}{dx}[\tan^{-1}(3x)] = \frac{1}{1+(3x)^2} \cdot 3 = \frac{3}{1+9x^2}$$

Level 2 Chain Rule with Square Root

Find $\frac{d}{dx}[\sin^{-1}(\sqrt{x})]$ for $0 < x < 1$.

Thought Process

Let $u = \sqrt{x} = x^{1/2}$.

Apply $\frac{d}{dx}(\sin^{-1}u) = \frac{1}{\sqrt{1-u^2}} \cdot \frac{du}{dx}$.

Remember $\frac{d}{dx}(\sqrt{x}) = \frac{1}{2\sqrt{x}}$.

Show Answer

$$\frac{d}{dx}[\sin^{-1}(\sqrt{x})] = \frac{1}{\sqrt{1-(\sqrt{x})^2}} \cdot \frac{1}{2\sqrt{x}}$$

$$= \frac{1}{\sqrt{1-x}} \cdot \frac{1}{2\sqrt{x}} = \frac{1}{2\sqrt{x}\sqrt{1-x}} = \frac{1}{2\sqrt{x(1-x)}}$$

Level 3 Product Rule Combination

Find $\frac{d}{dx}[x \cdot \arctan(x)]$.

Thought Process

This is a product of $x$ and $\arctan(x)$.

Apply the product rule: $(uv)' = u'v + uv'$.

Let $u = x$ (so $u' = 1$) and $v = \arctan(x)$ (so $v' = \frac{1}{1+x^2}$).

Show Answer

Using the product rule:

$$\frac{d}{dx}[x \cdot \arctan(x)] = 1 \cdot \arctan(x) + x \cdot \frac{1}{1+x^2}$$

$$= \arctan(x) + \frac{x}{1+x^2}$$

Level 4 Chain Rule and Optimization

(a) Find $\frac{d}{dx}[\arctan(e^x)]$ and simplify.

(b) Determine the value of $x$ where this derivative is maximized, and find the maximum value.

Thought Process

Part (a): Apply the chain rule with $u = e^x$.

The derivative of $\arctan(u)$ is $\frac{1}{1+u^2}$, and multiply by $\frac{du}{dx} = e^x$.

Part (b): To find the maximum of $f'(x)$, we need to find where $f''(x) = 0$.

Use the quotient rule to differentiate $f'(x) = \frac{e^x}{1+e^{2x}}$.

Show Answer

(a) Finding the derivative:

Let $u = e^x$, so $\frac{du}{dx} = e^x$.

$$\frac{d}{dx}[\arctan(e^x)] = \frac{1}{1+(e^x)^2} \cdot e^x = \frac{e^x}{1+e^{2x}}$$

(b) Finding the maximum:

Let $f(x) = \arctan(e^x)$, so $f'(x) = \frac{e^x}{1+e^{2x}}$.

Using the quotient rule: $$f''(x) = \frac{e^x(1+e^{2x}) - e^x(2e^{2x})}{(1+e^{2x})^2} = \frac{e^x(1-e^{2x})}{(1+e^{2x})^2}$$

Setting $f''(x) = 0$: Since $e^x > 0$ and $(1+e^{2x})^2 > 0$, we need $1 - e^{2x} = 0$.

Solving: $e^{2x} = 1 \Rightarrow 2x = 0 \Rightarrow x = 0$.

The maximum value of the derivative is: $$f'(0) = \frac{e^0}{1+e^0} = \frac{1}{1+1} = \frac{1}{2}$$

Interpretation: The function $\arctan(e^x)$ increases most rapidly at $x = 0$, where its slope is $1/2$.

Level 5 Prove an Identity Using Calculus

Use calculus to prove that $\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}$ for all $x \in [-1, 1]$.

Hint: Show the derivative of the left side is zero, then evaluate at a specific point.

Thought Process

Let $f(x) = \sin^{-1}x + \cos^{-1}x$.

If we can show $f'(x) = 0$ for all $x$ in the domain, then $f(x)$ is constant.

To find the constant, evaluate $f$ at an easy point like $x = 0$.

Show Answer

Step 1: Define $f(x) = \sin^{-1}x + \cos^{-1}x$ for $x \in [-1, 1]$.

Step 2: Compute the derivative: $$f'(x) = \frac{1}{\sqrt{1-x^2}} + \left(-\frac{1}{\sqrt{1-x^2}}\right) = 0$$

Step 3: Since $f'(x) = 0$ for all $x \in (-1, 1)$, the function $f(x)$ is constant.

Step 4: Find the constant by evaluating at $x = 0$: $$f(0) = \sin^{-1}(0) + \cos^{-1}(0) = 0 + \frac{\pi}{2} = \frac{\pi}{2}$$

Conclusion: For all $x \in [-1, 1]$: $$\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}$$

This elegant proof shows how calculus can establish identities: if a function’s derivative is zero everywhere on an interval, the function must be constant.

Conceptual Check (CCI-Style)

Question: The derivative of $\sin^{-1}(x)$ is $\frac{1}{\sqrt{1-x^2}}$. Why is this derivative undefined at $x = \pm 1$?

(A) Because $\sin^{-1}(\pm 1)$ doesn’t exist (B) Because the graph of $\sin^{-1}(x)$ has vertical tangent lines at $x = \pm 1$ (C) Because you can’t take the derivative of inverse functions at endpoints (D) Because $\sqrt{1-x^2}$ is complex-valued at $x = \pm 1$

Answer

(B) At $x = 1$, the graph of $\sin^{-1}(x)$ reaches its maximum value $\pi/2$, and the tangent line there is vertical (infinite slope). Similarly at $x = -1$, the tangent is vertical.

This makes geometric sense: as $x$ approaches $\pm 1$, the arcsine curve “flattens out” in the $x$-direction but continues to rise in the $y$-direction, creating a vertical tangent.

Note: Option (A) is wrong because $\sin^{-1}(\pm 1)$ does exist (they equal $\pm\pi/2$). Option (C) is a misstatement; you can often differentiate at endpoints. Option (D) is wrong because $\sqrt{0} = 0$, not complex, but the issue is that zero is in the denominator.

Common Pitfalls

Mistake	Why It’s Wrong	Correct Approach
Forgetting the negative sign for $\cos^{-1}$	$\frac{d}{dx}\cos^{-1}x = -\frac{1}{\sqrt{1-x^2}}$, not positive	Remember: $\sin^{-1}x + \cos^{-1}x = \pi/2$, so their derivatives must be negatives
Forgetting to apply chain rule	$\frac{d}{dx}\sin^{-1}(2x) \ne \frac{1}{\sqrt{1-4x^2}}$	Must include $\frac{d}{dx}(2x) = 2$: answer is $\frac{2}{\sqrt{1-4x^2}}$
Using wrong formula for $\sec^{-1}$	Some textbooks use different ranges for $\sec^{-1}$	Stewart uses $\frac{d}{dx}\sec^{-1}x = \frac{1}{\vert x\vert \sqrt{x^2-1}}$ with the absolute value
Thinking the derivative is only valid at interior points	The formula fails at $x = \pm 1$ for arcsine/arccosine	Domain of the derivative is $(-1, 1)$, not $[-1, 1]$

Common Misconceptions

Common misconception

the derivative of $\arcsin(g(x))$ is $\frac{1}{\sqrt{1-g(x)^2}}$, without the chain rule factor $g'(x)$.

This is the composition-is-not-chaining error. The chain rule requires $\frac{d}{dx}[\arcsin(g(x))] = \frac{g'(x)}{\sqrt{1 - [g(x)]^2}}$. For $y = \arcsin(x^2)$, the derivative is $\frac{2x}{\sqrt{1-x^4}}$, not $\frac{1}{\sqrt{1-x^4}}$. The factor $g'(x) = 2x$ must be included because the outer inverse-sine function is composed with the inner function $x^2$.

Mastery Checklist

I can state the derivatives of $\sin^{-1}x$, $\cos^{-1}x$, and $\tan^{-1}x$
I can derive these formulas using implicit differentiation
I can apply the chain rule to inverse trig compositions
I can simplify expressions like $\cos(\tan^{-1}x)$ using right triangles
I understand why the domains exclude certain values (division by zero in the formula)

Mental Model

Right Triangle Visualization:

When you see $\tan^{-1}x$, imagine a right triangle where:

One leg has length 1 (adjacent)
The other leg has length $x$ (opposite)
The hypotenuse has length $\sqrt{1+x^2}$

This triangle lets you convert any expression involving $\sin(\tan^{-1}x)$, $\cos(\tan^{-1}x)$, etc. into algebraic form without memorizing formulas.

                    /|
                   / |
    √(1+x²)       /  | x
                 /   |
                /θ   |
               ──────┘
                  1

tan θ = x/1 = x
sin θ = x/√(1+x²)
cos θ = 1/√(1+x²)

Connections

Looking back:

Implicit differentiation is how we derive these formulas
Trig identities simplify the derivations

Looking ahead:

These derivatives lead to integration formulas: $\int\frac{1}{\sqrt{1-x^2}}\,dx = \sin^{-1}x + C$
Trig substitution in integration uses these relationships in reverse

Real-world connections:

In physics, $\tan^{-1}$ appears in calculating angles from coordinates
In engineering, inverse trig functions describe phase angles in AC circuits

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Last updated: 2026-01-22