# Inverse Trig Derivatives Practice Problems with Answers

> 12 inverse trig derivative practice problems with answers and full solutions: arcsin, arccos, arctan and arcsec with the chain and product rules.

Source: https://practicecalculus.com/practice-problems/inverse-trig-derivatives
Updated: 2026-11-06

$$ \frac{d}{dx}\arcsin x = \frac{1}{\sqrt{1 - x^2}}, \qquad \frac{d}{dx}\arctan x = \frac{1}{1 + x^2} $$

## At a glance

- **Problems:** 12, with answers and full solutions
- **Level:** Calculus 1 and AP Calculus AB, Unit 3
- **You need:** The [chain rule](/practice-problems/chain-rule) and the four inverse trig derivatives
- **Main idea:** Use the formula with u in place of x, then multiply by u'
- **Practice with new numbers:** [Inverse trig derivatives lesson](/learn/calculus-1/derivative-rules/inverse-trig-derivatives)

## Key takeaways

- Arcsin and arccos have derivatives that differ only by a sign. The same is true for arctan and arccot.
- With an inside function, replace $x$ with $u$ everywhere in the formula, including under the root, then multiply by $u'$.
- If a derivative comes out as $0$, the function is constant. Problem 11 shows $\arctan x + \arctan\frac{1}{x} = \frac{\pi}{2}$ for $x > 0$.

These **inverse trig derivatives practice problems** start with the basic formulas, then use the chain rule, the product rule and a log, and finish with two problems that simplify in a surprising way and a tangent line. Try each one first. The answers come right after the list, then the full solutions.

The formulas you need:

| Function | Derivative |
| --- | --- |
| $\arcsin x$ | $\dfrac{1}{\sqrt{1 - x^2}}$ |
| $\arccos x$ | $-\dfrac{1}{\sqrt{1 - x^2}}$ |
| $\arctan x$ | $\dfrac{1}{1 + x^2}$ |
| $\operatorname{arcsec} x$ | $\dfrac{1}{\lvert x\rvert\sqrt{x^2 - 1}}$ |

Each formula comes from [implicit differentiation](/practice-problems/implicit-differentiation). See the [derivative of arcsin x](/derivatives/arcsin-x), the [derivative of arctan x](/derivatives/arctan-x) and the [derivative of arcsec x](/derivatives/arcsec-x) for the full proofs.

## Problems

**Basic formulas**

1. $y = 3\arctan x - 2\arcsin x$
2. $y = \arctan(3x)$
3. $y = \arccos(2x)$

**Chain rule**

4. $y = \arcsin(x^2)$
5. $y = \arctan\!\left(\sqrt{x}\right)$
6. $y = \arctan(e^x)$
7. $y = \operatorname{arcsec}(3x)$ for $x > \frac{1}{3}$

**With other rules**

8. $y = x\arctan x$
9. $y = \ln(\arctan x)$ for $x > 0$
10. $y = \arcsin x + \sqrt{1 - x^2}$

**Simplify and apply**

11. $y = \arctan x + \arctan\!\left(\dfrac{1}{x}\right)$ for $x > 0$
12. Find the tangent line to $y = \arctan x$ at $x = 1$.

## Answers

| Problem | Answer |
| --- | --- |
| 1 | $\dfrac{3}{1 + x^2} - \dfrac{2}{\sqrt{1 - x^2}}$ |
| 2 | $\dfrac{3}{1 + 9x^2}$ |
| 3 | $-\dfrac{2}{\sqrt{1 - 4x^2}}$ |
| 4 | $\dfrac{2x}{\sqrt{1 - x^4}}$ |
| 5 | $\dfrac{1}{2\sqrt{x}\,(1 + x)}$ |
| 6 | $\dfrac{e^x}{1 + e^{2x}}$ |
| 7 | $\dfrac{1}{x\sqrt{9x^2 - 1}}$ |
| 8 | $\arctan x + \dfrac{x}{1 + x^2}$ |
| 9 | $\dfrac{1}{(1 + x^2)\arctan x}$ |
| 10 | $\dfrac{1 - x}{\sqrt{1 - x^2}} = \sqrt{\dfrac{1 - x}{1 + x}}$ |
| 11 | $0$ |
| 12 | $y = \dfrac{1}{2}x - \dfrac{1}{2} + \dfrac{\pi}{4}$ |

## Step-by-step solutions

**1.** Use the table for each term:

$$y' = \frac{3}{1 + x^2} - \frac{2}{\sqrt{1 - x^2}}$$

**2.** With $u = 3x$, $u' = 3$ and $u^2 = 9x^2$:

$$y' = \frac{3}{1 + (3x)^2} = \frac{3}{1 + 9x^2}$$

**3.** With $u = 2x$, $u' = 2$ and $u^2 = 4x^2$. Keep the minus sign from arccos:

$$y' = -\frac{2}{\sqrt{1 - 4x^2}}$$

**4.** With $u = x^2$, $u' = 2x$ and $u^2 = x^4$:

$$y' = \frac{2x}{\sqrt{1 - x^4}}$$

**5.** With $u = \sqrt{x}$, $u' = \frac{1}{2\sqrt{x}}$ and $u^2 = x$:

$$y' = \frac{1}{1 + x}\cdot\frac{1}{2\sqrt{x}} = \frac{1}{2\sqrt{x}\,(1 + x)}$$

**6.** With $u = e^x$, $u' = e^x$ and $u^2 = e^{2x}$:

$$y' = \frac{e^x}{1 + e^{2x}}$$

**7.** With $u = 3x > 1$, $|u| = 3x$ and $u' = 3$:

$$y' = \frac{3}{3x\sqrt{9x^2 - 1}} = \frac{1}{x\sqrt{9x^2 - 1}}$$

**8.** Use the [product rule](/rules/product-rule):

$$y' = 1\cdot\arctan x + x\cdot\frac{1}{1 + x^2} = \arctan x + \frac{x}{1 + x^2}$$

**9.** Use $(\ln u)' = \frac{u'}{u}$ with $u = \arctan x$:

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

**10.** The derivative of $\sqrt{1 - x^2}$ is $\frac{-2x}{2\sqrt{1 - x^2}} = -\frac{x}{\sqrt{1 - x^2}}$. Add the two parts:

$$y' = \frac{1}{\sqrt{1 - x^2}} - \frac{x}{\sqrt{1 - x^2}} = \frac{1 - x}{\sqrt{1 - x^2}}$$

Since $1 - x^2 = (1 - x)(1 + x)$, this simplifies to $\sqrt{\frac{1 - x}{1 + x}}$ for $-1 < x < 1$.

**11.** For the second term, $u = \frac{1}{x}$ and $u' = -\frac{1}{x^2}$:

$$\frac{d}{dx}\arctan\frac{1}{x} = \frac{-\frac{1}{x^2}}{1 + \frac{1}{x^2}} = -\frac{1}{x^2 + 1}$$

So $y' = \frac{1}{1 + x^2} - \frac{1}{1 + x^2} = 0$. The function is constant for $x > 0$. At $x = 1$ it equals $\frac{\pi}{4} + \frac{\pi}{4} = \frac{\pi}{2}$, so $\arctan x + \arctan\frac{1}{x} = \frac{\pi}{2}$ for every $x > 0$.

**12.** The point is $\arctan 1 = \frac{\pi}{4}$. The slope is $\frac{1}{1 + 1^2} = \frac{1}{2}$:

$$y - \frac{\pi}{4} = \frac{1}{2}(x - 1) \quad\Rightarrow\quad y = \frac{1}{2}x - \frac{1}{2} + \frac{\pi}{4}$$

## Common mistakes on these problems

- **Squaring only $x$, not the whole inside.** In problem 2 the bottom is $1 + (3x)^2 = 1 + 9x^2$, not $1 + 3x^2$.
- **Forgetting $u'$ on top.** $\frac{d}{dx}\arcsin(x^2)$ needs the $2x$ in the numerator.
- **Mixing up $\arcsin x$ and $\frac{1}{\sin x}$.** $\arcsin x$ is the inverse function, and $\frac{1}{\sin x}$ is $\csc x$. Their derivatives are completely different.
- **Dropping the minus on arccos.** Problem 3 has a negative answer because $(\arccos x)' = -\frac{1}{\sqrt{1 - x^2}}$.

## Keep practicing

These 12 problems use fixed numbers. For unlimited practice with new functions and instant feedback, try the [inverse trig derivatives lesson](/learn/calculus-1/derivative-rules/inverse-trig-derivatives) and the [inverse function derivatives lesson](/learn/calculus-1/derivative-rules/inverse-function-derivatives). To review the regular trig functions first, use the [trig derivatives practice problems](/practice-problems/derivatives-of-trig-functions).

## Frequently asked questions

### What are the derivatives of the inverse trig functions?

$(\arcsin x)' = \frac{1}{\sqrt{1-x^2}}$, $(\arccos x)' = -\frac{1}{\sqrt{1-x^2}}$, $(\arctan x)' = \frac{1}{1+x^2}$, $(\operatorname{arccot} x)' = -\frac{1}{1+x^2}$, $(\operatorname{arcsec} x)' = \frac{1}{|x|\sqrt{x^2-1}}$ and $(\operatorname{arccsc} x)' = -\frac{1}{|x|\sqrt{x^2-1}}$.

### How do you find the derivative of arcsin(u)?

Use the chain rule: $\frac{d}{dx}\arcsin u = \frac{u'}{\sqrt{1 - u^2}}$. For $\arcsin(x^2)$, $u = x^2$ and $u' = 2x$, so the derivative is $\frac{2x}{\sqrt{1 - x^4}}$.

### How is the derivative of arcsin x derived?

Write $y = \arcsin x$ as $\sin y = x$ and differentiate implicitly: $\cos y\,y' = 1$. Since $\cos y = \sqrt{1 - \sin^2 y} = \sqrt{1 - x^2}$ on the range of arcsin, $y' = \frac{1}{\sqrt{1 - x^2}}$.