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The derivatives of $\sin^{-1}x$ and $\cos^{-1}x$ have $\sqrt{1-x^2}$ in them. This isn't 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.
| Property | Value |
|---|---|
| Concept | Inverse Function Derivatives |
| Chapter | 3.6 |
| Difficulty | Intermediate |
| Time | ~25 minutes |
$$\boxed{\frac{d}{dx}(\sin^{-1}x) = \frac{1}{\sqrt{1-x^2}}, \quad -1 < x < 1}$$
$$\boxed{\frac{d}{dx}(\cos^{-1}x) = -\frac{1}{\sqrt{1-x^2}}, \quad -1 < x < 1}$$
$$\boxed{\frac{d}{dx}(\tan^{-1}x) = \frac{1}{1+x^2}, \quad x \in \mathbb{R}}$$
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: $$\frac{d}{dx}(\sin^{-1}x) = \frac{1}{\sqrt{1-x^2}}$$
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 $\sec^2 y = 1 + \tan^2 y$: $$\frac{dy}{dx} = \frac{1}{1 + \tan^2 y} = \frac{1}{1 + x^2}$$
For a differentiable function $u = g(x)$:
| Formula | Chain Rule Version |
|---|---|
| $\frac{d}{dx}(\sin^{-1}u)$ | $\frac{1}{\sqrt{1-u^2}} \cdot \frac{du}{dx}$ |
| $\frac{d}{dx}(\cos^{-1}u)$ | $-\frac{1}{\sqrt{1-u^2}} \cdot \frac{du}{dx}$ |
| $\frac{d}{dx}(\tan^{-1}u)$ | $\frac{1}{1+u^2} \cdot \frac{du}{dx}$ |
| 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⁻¹x domain: ←───|═══════|───→
-1 1
tan⁻¹x domain: ←═══════════════→
all real numbers
sec⁻¹x domain: ←═══| |═══→
-1 1
(excludes middle)Find $\frac{d}{dx}[\tan^{-1}(3x)]$.
Find $\frac{d}{dx}[\sin^{-1}(\sqrt{x})]$ for $0 < x < 1$.
Find $\frac{d}{dx}[x \cdot \arctan(x)]$.
(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.
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.
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 $x = \pm 1$
(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.
| 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 $[-1, 1]$ |
Right Triangle Visualization:
When you see $\tan^{-1}x$, imagine a right triangle where:
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)Looking back:
Looking ahead:
Real-world connections:
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|---|---|---|
| Logarithmic Differentiation | Skills Index | Integration Techniques |
Last updated: 2026-01-22