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.
Derivatives Updated 4 min read
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
- Main idea
- Learn the six basic derivatives, then combine them with the other rules
- Practice with new numbers
- Trig derivatives lesson
On this page
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 and the derivative of sec x. For the big picture, read the trig derivatives guide.
Problems
Basic derivatives
- y = 3\sin x - 2\cos x
- y = \tan x + \sec x
- y = 4\csc x - \cot x
Product and quotient rules
- y = x^2\cos x
- y = \sin x\cos x
- y = \dfrac{\sin x}{1 + \cos x}
Chain rule
- y = \sin(5x)
- y = \cos(x^2)
- y = \tan^3 x
- y = \sec(3x)
Second derivatives and tangent lines
- Find y'' for y = \sin(2x).
- 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:
2. Use the two derivatives straight from the table:
3. Both \csc x and \cot x have negative derivatives:
4. Use the product rule with f = x^2 and g = \cos x:
5. Product rule again:
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. The bottom's derivative is -\sin x:
Since \cos^2 x + \sin^2 x = 1, the top is 1 + \cos x, and one factor cancels:
7. The inside is 5x, with derivative 5:
8. The inside is x^2, with derivative 2x:
9. Write y = (\tan x)^3. The outside is a cube and the inside is \tan x:
10. The outside is \sec u with derivative \sec u\tan u, and u = 3x:
11. Differentiate twice, using the chain rule each time:
12. The point is \tan\frac{\pi}{4} = 1. The slope is \sec^2\frac{\pi}{4} = \left(\sqrt{2}\right)^2 = 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 and then the chain rule with trig lesson. For the inverse functions, like \arcsin x and \arctan x, move on to the inverse trig derivatives practice problems.
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