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Derivatives

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

\frac{d}{dx}\sin x = \cos x, \quad \frac{d}{dx}\cos x = -\sin x, \quad \frac{d}{dx}\tan x = \sec^2 x
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
  1. Problems
  2. Answers
  3. Step-by-step solutions
  4. Common mistakes on these problems
  5. Keep practicing

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

  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

  1. y = x^2\cos x
  2. y = \sin x\cos x
  3. y = \dfrac{\sin x}{1 + \cos x}

Chain rule

  1. y = \sin(5x)
  2. y = \cos(x^2)
  3. y = \tan^3 x
  4. y = \sec(3x)

Second derivatives and tangent lines

  1. Find y'' for y = \sin(2x).
  2. 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 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. 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 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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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.

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