Blog · Calculus guides · 30 October 2026 · 4 min read

By the Asymptote team. Every answer on this page is checked by the same machine that grades the game.

Derivatives of Trig Functions: Table & Practice Problems

An illustration of Derivatives of trig functions

The six trig derivatives are $\frac{d}{d x} sin x = cos x$ , $\frac{d}{d x} cos x = - sin x$ , $\frac{d}{d x} tan x = sec^{2} x$ , $\frac{d}{d x} cot x = - csc^{2} x$ , $\frac{d}{d x} sec x = sec x tan x$ and $\frac{d}{d x} csc x = - csc x cot x$ . The three that start with "co" all have a minus sign. Inside anything other than plain $x$ , multiply by the inside's derivative as well.

The formula

\frac{d}{d x} sin x = cos x \frac{d}{d x} tan x = sec^{2} x \frac{d}{d x} sec x = sec x tan x \frac{d}{d x} cos x = - sin x \frac{d}{d x} cot x = - csc^{2} x \frac{d}{d x} csc x = - csc x cot x

Read across each row: the right-hand column is the left-hand one with every function swapped for its "co" partner and a minus sign stuck on the front. Six facts for the price of three.

When to use it

Use these any time a trig function needs differentiating, which in practice means they are rarely used alone. A coefficient just comes along for the ride, a product like $x sec x$ needs the product rule, and $sin$ of anything other than $x$ needs the chain rule. The rules here only hold with $x$ in radians: in degrees, every derivative picks up a factor of $\frac{π}{180}$ , which is one of many reasons calculus quietly assumes radians. You only truly need two of the six from memory, sine and cosine; the other four are quotients of them, so the quotient rule rebuilds any one you forget in about thirty seconds.

How to do it, step by step

Learn sine and tangent cold. $sin \to cos$ and $tan \to sec^{2}$ . Secant is $sec x tan x$ , the only one that comes out as a product.
Get the other three by the "co" swap. Swap every function for its partner (sin and cos, tan and cot, sec and csc) and add a minus sign: $tan \to sec^{2}$ becomes $cot \to - csc^{2}$ .
Spot any product or composition. A trig function times something needs the product rule; a trig function of something needs the chain rule. So $tan (5 x)$ becomes $5 sec^{2} (5 x)$ , and $x cos x$ becomes $cos x - x sin x$ .
Simplify at the end, not the start. Factor out common pieces such as $sec x$ once the differentiating is done.

Worked examples

Example 1: Three different functions

Easy

Differentiate:

y = 4 sin x + 3 tan x - cot x

A sum differentiates term by term, and a constant multiple just waits at the front while you differentiate what it multiplies. So this is three table lookups and some care with signs.

\frac{d y}{d x} = 4 cos x + 3 sec^{2} x - (- csc^{2} x)

The derivative of $cot x$ is $- csc^{2} x$ , and subtracting a negative turns it positive. That double minus is exactly where these questions go wrong.

Answer

\frac{d y}{d x} = 4 cos x + 3 sec^{2} x + csc^{2} x

Example 2: Secant in a product

Medium

Differentiate:

y = x sec x

This is $u = x$ times $v = sec x$ , so use the product rule $u^{'} v + u v^{'}$ with $u^{'} = 1$ and $v^{'} = sec x tan x$ .

\frac{d y}{d x} = 1 \cdot sec x + x \cdot sec x tan x

If you like, factor out $sec x$ to get $sec x (1 + x tan x)$ . Both forms are the same answer, and both are accepted wherever answers are checked as maths rather than as strings.

If you would rather not memorize $sec x tan x$ , write $sec x = \frac{1}{c o s x}$ and use the chain rule: $- \frac{1}{c o s ^{2} x} \cdot (- sin x) = \frac{s i n x}{c o s ^{2} x}$ , which is $\frac{1}{c o s x} \cdot \frac{s i n x}{c o s x}$ . Same thing, rebuilt from scratch.

Answer

\frac{d y}{d x} = sec x + x sec x tan x

Example 3: A logarithm that collapses

Harder

Differentiate:

y = ln (sec (x) + tan (x))

The outside is $ln u$ , whose derivative is $\frac{1}{u}$ . The inside is $u = sec x + tan x$ , whose derivative is $sec x tan x + sec^{2} x$ .

\frac{d y}{d x} = \frac{sec x tan x + sec ^{2} x}{sec x + tan x}

The top has a common factor of $sec x$ : it is $sec x (tan x + sec x)$ . The bracket matches the bottom exactly, so it cancels.

\frac{d y}{d x} = \frac{sec x ( sec x + tan x )}{sec x + tan x} = sec x

So a function that looked like it would need three rules at once differentiates to a single term. When an answer simplifies this dramatically, it is usually a sign the question was built backwards from it.

Which is why the integral of $sec x$ is $ln ∣ sec x + tan x ∣ + C$ : a fact nobody guesses and everybody eventually memorizes.

Answer

\frac{d y}{d x} = sec x

Derivatives of trig functions practice problems

Three to try, easy to harder. Type an answer and it is checked instantly by the same checker that grades the game; any correct form counts. Stuck? The answer and the first step are under each one.

Problem 1

Easy

Differentiate:

y = 3 cos x - 2 sec x

Stuck? Show the answer

Answer

- 3 sin x - 2 sec x tan x

How to start

Term by term: $cos x$ gives $- sin x$ and $sec x$ gives $sec x tan x$ . Keep the coefficients and watch the signs.

Problem 2

Medium

Differentiate:

y = 4 sin (3 x^{2} - x)

Stuck? Show the answer

Answer

4 (6 x - 1) cos (3 x^{2} - x)

How to start

Chain rule: $sin u$ becomes $cos u$ with the inside left alone, then multiply by the inside's derivative, $6 x - 1$ .

Problem 3

Harder

Differentiate, and simplify:

y = \frac{sin x}{1 + cos x}

Stuck? Show the answer

Answer

\frac{1}{1 + cos x}

How to start

Quotient rule. The top comes out as $cos x (1 + cos x) + sin^{2} x$ ; use $sin^{2} x + cos^{2} x = 1$ and look for a factor that cancels with the bottom.

Want more? Asymptote has 1,060 more derivatives of trig functions problems, free, with no account.

Common mistakes

Losing the minus on cosine. $\frac{d}{d x} cos x = - sin x$ . Going round the cycle $sin \to cos \to - sin \to - cos \to sin$ helps: every second step changes sign.
Writing $sec x$ as the derivative of $tan x$ . It is $sec^{2} x$ , squared. In the other direction, $sec x$ differentiates to $sec x tan x$ , not $sec^{2} x$ .
Dropping the chain rule. $\frac{d}{d x} sin (3 x) = 3 cos (3 x)$ , not $cos (3 x)$ . In our answer data, plain standard derivatives like $\frac{d}{d x} sin x$ are right 93% of the time; with a chain rule wrapped round them it drops to 90%.
Squaring the wrong thing. $sin^{2} x$ means $(sin x)^{2}$ , so its derivative is $2 sin x cos x$ by the chain rule. It is not $sin (x^{2})$ , whose derivative is $2 x cos (x^{2})$ .
Working in degrees. All six formulas assume radians. A calculator left in degree mode will give answers that are wrong by a factor of $\frac{π}{180}$ .

Practice it now

Five minutes, no account

A Sprint can open with a trig derivative: single standard functions first, then sums of $sin$ , $cos$ , $tan$ and $sec$ , then trig of polynomials once the chain rule arrives. There are no $cot$ or $csc$ problems in the game yet, so the other three you will have to trust us on.

Play a Sprint See it on the formula sheet

FAQ

What is the derivative of $csc x$ ?: $- csc x cot x$ . It is the "co" version of $sec x \to sec x tan x$ : swap each function for its partner and add a minus sign.
Why is the derivative of $tan x$ equal to $sec^{2} x$ ?: Write $tan x = \frac{s i n x}{c o s x}$ and use the quotient rule: the top becomes $cos^{2} x + sin^{2} x = 1$ , leaving $\frac{1}{c o s ^{2} x} = sec^{2} x$ .
Which trig derivatives are negative?: The three co-functions: cosine, cotangent and cosecant. Sine, tangent and secant come out positive.
What is the second derivative of $sin x$ ?: $- sin x$ . Sine and cosine cycle with period four: $sin x \to cos x \to - sin x \to - cos x \to sin x$ . So the fourth derivative of $sin x$ is $sin x$ again, and the 100th is too.
Do trig derivatives work in degrees?: Not as written. The formula $\frac{d}{d x} sin x = cos x$ relies on $lim_{x \to 0} \frac{s i n x}{x} = 1$ , which is only true in radians.
Where can I find derivatives of trig functions practice problems?: Right on this page: the practice problems section has three derivatives of trig functions problems, easy, medium and harder, with the answer and the first step under each one, and a box that checks what you type with the same checker that grades Asymptote. The game has 1,060 more derivatives of trig functions problems after that, free and with no account: try a five-minute Sprint.

When to use it

Use these any time a trig function needs differentiating, which in practice means they are rarely used alone. A coefficient just comes along for the ride, a product like

x sec x

needs the product rule, and

sin

of anything other than

x

needs the chain rule. The rules here only hold with

x

in radians: in degrees, every derivative picks up a factor of

\frac{π}{180}

, which is one of many reasons calculus quietly assumes radians. You only truly need two of the six from memory, sine and cosine; the other four are quotients of them, so the quotient rule rebuilds any one you forget in about thirty seconds.

How to do it, step by step

Learn sine and tangent cold.

sin \to cos

and

tan \to sec^{2}

. Secant is

sec x tan x

, the only one that comes out as a product.

Get the other three by the "co" swap. Swap every function for its partner (sin and cos, tan and cot, sec and csc) and add a minus sign:

tan \to sec^{2}

becomes

cot \to - csc^{2}

Spot any product or composition. A trig function times something needs the product rule; a trig function of something needs the chain rule. So

tan (5 x)

becomes

5 sec^{2} (5 x)

, and

x cos x

becomes

cos x - x sin x

Simplify at the end, not the start. Factor out common pieces such as

sec x

once the differentiating is done.

Worked examples

Example 1: Three different functions

Easy

Differentiate:

y = 4 sin x + 3 tan x - cot x

A sum differentiates term by term, and a constant multiple just waits at the front while you differentiate what it multiplies. So this is three table lookups and some care with signs.

\frac{d y}{d x} = 4 cos x + 3 sec^{2} x - (- csc^{2} x)

The derivative of $cot x$ is $- csc^{2} x$ , and subtracting a negative turns it positive. That double minus is exactly where these questions go wrong.

Answer

\frac{d y}{d x} = 4 cos x + 3 sec^{2} x + csc^{2} x

Example 2: Secant in a product

Medium

Differentiate:

y = x sec x

This is $u = x$ times $v = sec x$ , so use the product rule $u^{'} v + u v^{'}$ with $u^{'} = 1$ and $v^{'} = sec x tan x$ .

\frac{d y}{d x} = 1 \cdot sec x + x \cdot sec x tan x

If you like, factor out $sec x$ to get $sec x (1 + x tan x)$ . Both forms are the same answer, and both are accepted wherever answers are checked as maths rather than as strings.

Answer

\frac{d y}{d x} = sec x + x sec x tan x

Example 3: A logarithm that collapses

Harder

Differentiate:

y = ln (sec (x) + tan (x))

The outside is $ln u$ , whose derivative is $\frac{1}{u}$ . The inside is $u = sec x + tan x$ , whose derivative is $sec x tan x + sec^{2} x$ .

\frac{d y}{d x} = \frac{sec x tan x + sec ^{2} x}{sec x + tan x}

The top has a common factor of $sec x$ : it is $sec x (tan x + sec x)$ . The bracket matches the bottom exactly, so it cancels.

\frac{d y}{d x} = \frac{sec x ( sec x + tan x )}{sec x + tan x} = sec x

Which is why the integral of $sec x$ is $ln ∣ sec x + tan x ∣ + C$ : a fact nobody guesses and everybody eventually memorizes.

Answer

\frac{d y}{d x} = sec x

Derivatives of trig functions practice problems

Three to try, easy to harder. Type an answer and it is checked instantly by the same checker that grades the game; any correct form counts. Stuck? The answer and the first step are under each one.

Problem 1

Easy

Differentiate:

y = 3 cos x - 2 sec x

Stuck? Show the answer

Answer

- 3 sin x - 2 sec x tan x

How to start

Term by term: $cos x$ gives $- sin x$ and $sec x$ gives $sec x tan x$ . Keep the coefficients and watch the signs.

Problem 2

Medium

Differentiate:

y = 4 sin (3 x^{2} - x)

Stuck? Show the answer

Answer

4 (6 x - 1) cos (3 x^{2} - x)

How to start

Chain rule: $sin u$ becomes $cos u$ with the inside left alone, then multiply by the inside's derivative, $6 x - 1$ .

Problem 3

Harder

Differentiate, and simplify:

y = \frac{sin x}{1 + cos x}

Stuck? Show the answer

Answer

\frac{1}{1 + cos x}

How to start

Quotient rule. The top comes out as $cos x (1 + cos x) + sin^{2} x$ ; use $sin^{2} x + cos^{2} x = 1$ and look for a factor that cancels with the bottom.

Common mistakes

Losing the minus on cosine.

\frac{d}{d x} cos x = - sin x

. Going round the cycle

sin \to cos \to - sin \to - cos \to sin

helps: every second step changes sign.

Writing $sec x$ as the derivative of $tan x$ . It is

sec^{2} x

, squared. In the other direction,

sec x

differentiates to

sec x tan x

, not

sec^{2} x

Dropping the chain rule.

\frac{d}{d x} sin (3 x) = 3 cos (3 x)

, not

cos (3 x)

. In our answer data, plain standard derivatives like

\frac{d}{d x} sin x

are right 93% of the time; with a chain rule wrapped round them it drops to 90%.

Squaring the wrong thing.

sin^{2} x

means

(sin x)^{2}

, so its derivative is

2 sin x cos x

by the chain rule. It is not

sin (x^{2})

, whose derivative is

2 x cos (x^{2})

Working in degrees. All six formulas assume radians. A calculator left in degree mode will give answers that are wrong by a factor of

\frac{π}{180}

Practice it now

Five minutes, no account

Play a Sprint See it on the formula sheet

FAQ

What is the derivative of

csc x

- csc x cot x

. It is the "co" version of

sec x \to sec x tan x

: swap each function for its partner and add a minus sign.

Why is the derivative of

tan x

equal to

sec^{2} x

Write

tan x = \frac{s i n x}{c o s x}

and use the quotient rule: the top becomes

cos^{2} x + sin^{2} x = 1

, leaving

\frac{1}{c o s ^{2} x} = sec^{2} x

Which trig derivatives are negative?

The three co-functions: cosine, cotangent and cosecant. Sine, tangent and secant come out positive.

What is the second derivative of

sin x

- sin x

. Sine and cosine cycle with period four:

sin x \to cos x \to - sin x \to - cos x \to sin x

. So the fourth derivative of

sin x

sin x

again, and the 100th is too.

Do trig derivatives work in degrees?

Not as written. The formula

\frac{d}{d x} sin x = cos x

relies on

lim_{x \to 0} \frac{s i n x}{x} = 1

, which is only true in radians.

Where can I find derivatives of trig functions practice problems?

Right on this page: the practice problems section has three derivatives of trig functions problems, easy, medium and harder, with the answer and the first step under each one, and a box that checks what you type with the same checker that grades Asymptote. The game has 1,060 more derivatives of trig functions problems after that, free and with no account: try a five-minute Sprint.

Derivatives of Trig Functions: Table & Practice Problems

The formula

When to use it

How to do it, step by step

Worked examples

Example 1: Three different functions

Example 2: Secant in a product

Example 3: A logarithm that collapses

Derivatives of trig functions practice problems

Common mistakes

FAQ

Keep going

Derivatives of Trig Functions: Table & Practice Problems

The formula

When to use it

How to do it, step by step

Worked examples

Example 1: Three different functions

Example 2: Secant in a product

Example 3: A logarithm that collapses

Derivatives of trig functions practice problems

Common mistakes

FAQ

Keep going