# Trig Derivatives Practice Problems with Answers and Solutions

> 12 derivatives of trig functions practice problems with answers and full solutions: sin, cos, tan, sec, csc, cot, with product, quotient and chain rules.

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

$$ \frac{d}{dx}\sin x = \cos x, \quad \frac{d}{dx}\cos x = -\sin x, \quad \frac{d}{dx}\tan x = \sec^2 x $$

## At a glance

- **Problems:** 12, with answers and full solutions
- **Level:** Calculus 1 and AP Calculus AB, Unit 2 and Unit 3
- **You need:** The six trig derivatives and the [product and quotient rules](/practice-problems/product-and-quotient-rules)
- **Main idea:** Learn the six basic derivatives, then combine them with the other rules
- **Practice with new numbers:** [Trig derivatives lesson](/learn/calculus-1/derivative-rules/trig-derivatives)

## Key takeaways

- The derivatives of the "co" functions, $\cos x$, $\csc x$ and $\cot x$, all start with a minus sign.
- Pairs to remember: $\tan x \to \sec^2 x$, $\sec x \to \sec x\tan x$, $\cot x \to -\csc^2 x$, $\csc x \to -\csc x\cot x$.
- After a quotient rule, look for $\sin^2 x + \cos^2 x = 1$. It often collapses the answer, as in problem 6.

These **derivatives of trig functions practice problems** start with the six basic derivatives, add the product and quotient rules, then the chain rule, and finish with a second derivative and a tangent line. Try each one first. The answers come right after the list, then the full solutions.

The six derivatives you need:

| Function | Derivative |
| --- | --- |
| $\sin x$ | $\cos x$ |
| $\cos x$ | $-\sin x$ |
| $\tan x$ | $\sec^2 x$ |
| $\cot x$ | $-\csc^2 x$ |
| $\sec x$ | $\sec x\tan x$ |
| $\csc x$ | $-\csc x\cot x$ |

Each one has its own page with a proof, for example the [derivative of tan x](/derivatives/tan-x) and the [derivative of sec x](/derivatives/sec-x). For the big picture, read the [trig derivatives guide](/blog/derivatives-of-trig-functions).

## Problems

**Basic derivatives**

1. $y = 3\sin x - 2\cos x$
2. $y = \tan x + \sec x$
3. $y = 4\csc x - \cot x$

**Product and quotient rules**

4. $y = x^2\cos x$
5. $y = \sin x\cos x$
6. $y = \dfrac{\sin x}{1 + \cos x}$

**Chain rule**

7. $y = \sin(5x)$
8. $y = \cos(x^2)$
9. $y = \tan^3 x$
10. $y = \sec(3x)$

**Second derivatives and tangent lines**

11. Find $y''$ for $y = \sin(2x)$.
12. Find the tangent line to $y = \tan x$ at $x = \frac{\pi}{4}$.

## Answers

| Problem | Answer |
| --- | --- |
| 1 | $3\cos x + 2\sin x$ |
| 2 | $\sec^2 x + \sec x\tan x$ |
| 3 | $-4\csc x\cot x + \csc^2 x$ |
| 4 | $2x\cos x - x^2\sin x$ |
| 5 | $\cos^2 x - \sin^2 x = \cos(2x)$ |
| 6 | $\dfrac{1}{1 + \cos x}$ |
| 7 | $5\cos(5x)$ |
| 8 | $-2x\sin(x^2)$ |
| 9 | $3\tan^2 x\sec^2 x$ |
| 10 | $3\sec(3x)\tan(3x)$ |
| 11 | $-4\sin(2x)$ |
| 12 | $y = 2x + 1 - \dfrac{\pi}{2}$ |

## Step-by-step solutions

**1.** Differentiate each term. The minus from $(\cos x)' = -\sin x$ meets the $-2$:

$$y' = 3\cos x - 2(-\sin x) = 3\cos x + 2\sin x$$

**2.** Use the two derivatives straight from the table:

$$y' = \sec^2 x + \sec x\tan x$$

**3.** Both $\csc x$ and $\cot x$ have negative derivatives:

$$y' = 4(-\csc x\cot x) - (-\csc^2 x) = -4\csc x\cot x + \csc^2 x$$

**4.** Use the [product rule](/rules/product-rule) with $f = x^2$ and $g = \cos x$:

$$y' = 2x\cos x + x^2(-\sin x) = 2x\cos x - x^2\sin x$$

**5.** Product rule again:

$$y' = \cos x\cdot\cos x + \sin x\cdot(-\sin x) = \cos^2 x - \sin^2 x$$

By the double-angle identity this equals $\cos(2x)$. You can confirm it another way: $\sin x\cos x = \frac{1}{2}\sin(2x)$, whose derivative is $\cos(2x)$.

**6.** Use the [quotient rule](/rules/quotient-rule). The bottom's derivative is $-\sin x$:

$$y' = \frac{\cos x(1 + \cos x) - \sin x(-\sin x)}{(1 + \cos x)^2} = \frac{\cos x + \cos^2 x + \sin^2 x}{(1 + \cos x)^2}$$

Since $\cos^2 x + \sin^2 x = 1$, the top is $1 + \cos x$, and one factor cancels:

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

**7.** The inside is $5x$, with derivative $5$:

$$y' = \cos(5x)\cdot 5 = 5\cos(5x)$$

**8.** The inside is $x^2$, with derivative $2x$:

$$y' = -\sin(x^2)\cdot 2x = -2x\sin(x^2)$$

**9.** Write $y = (\tan x)^3$. The outside is a cube and the inside is $\tan x$:

$$y' = 3(\tan x)^2\cdot\sec^2 x = 3\tan^2 x\sec^2 x$$

**10.** The outside is $\sec u$ with derivative $\sec u\tan u$, and $u = 3x$:

$$y' = \sec(3x)\tan(3x)\cdot 3 = 3\sec(3x)\tan(3x)$$

**11.** Differentiate twice, using the chain rule each time:

$$y' = 2\cos(2x), \qquad y'' = 2\cdot(-\sin(2x))\cdot 2 = -4\sin(2x)$$

**12.** The point is $\tan\frac{\pi}{4} = 1$. The slope is $\sec^2\frac{\pi}{4} = \left(\sqrt{2}\right)^2 = 2$:

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

## Common mistakes on these problems

- **Dropping the minus sign.** The derivatives of $\cos x$, $\cot x$ and $\csc x$ are all negative. Problem 3 has two of them.
- **Writing $\sec^2 x$ as $\sec(x^2)$.** $\sec^2 x$ means $(\sec x)^2$. These are very different functions.
- **Forgetting the inside derivative.** $\sin(5x)$ has derivative $5\cos(5x)$, not $\cos(5x)$.
- **Using degrees.** In problem 12 the angle is $\frac{\pi}{4}$ radians. Using $45$ in the tangent line would be wrong.

## Keep practicing

These 12 problems use fixed numbers. For unlimited practice with new functions and instant feedback, try the [trig derivatives lesson](/learn/calculus-1/derivative-rules/trig-derivatives) and then the [chain rule with trig lesson](/learn/calculus-1/derivative-rules/chain-rule-trig). For the inverse functions, like $\arcsin x$ and $\arctan x$, move on to the [inverse trig derivatives practice problems](/practice-problems/inverse-trig-derivatives).

## Frequently asked questions

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

$(\sin x)' = \cos x$, $(\cos x)' = -\sin x$, $(\tan x)' = \sec^2 x$, $(\cot x)' = -\csc^2 x$, $(\sec x)' = \sec x\tan x$ and $(\csc x)' = -\csc x\cot x$. All of these use radians.

### Why is the derivative of sin x equal to cos x?

From the limit definition, $\frac{\sin(x+h) - \sin x}{h}$ splits with the angle sum formula into $\sin x\cdot\frac{\cos h - 1}{h} + \cos x\cdot\frac{\sin h}{h}$. As $h\to 0$ the first limit is $0$ and the second is $1$, which leaves $\cos x$.

### Do trig derivatives work in degrees?

Not in this form. The rule $(\sin x)' = \cos x$ needs $x$ in radians. In degrees an extra factor of $\frac{\pi}{180}$ appears, which is why calculus always uses radians.