Calculus — Derivatives⏱ 6 min read

Trigonometric Derivatives

d/dx(sin x) = cos x

What is the Trigonometric Derivatives?

These are the derivatives of all six basic trigonometric functions, essential to memorize for calculus: d/dx(sin x) = cos x, d/dx(cos x) = −sin x, d/dx(tan x) = sec²x, d/dx(cot x) = −csc²x, d/dx(sec x) = sec x·tan x, and d/dx(csc x) = −csc x·cot x.

A useful pattern: sine and cosine are each other's derivatives (with a sign flip going from cosine to sine), and every "co-" function (cos, cot, csc) has a negative derivative.

What Each Variable Means

sec x

SecantEqual to 1/cos x — the reciprocal of cosine.

csc x

CosecantEqual to 1/sin x — the reciprocal of sine.

cot x

CotangentEqual to cos x/sin x — the reciprocal of tangent.

When to Use It

Differentiating any expression built from trigonometric functions
Combined with the chain, product, and quotient rules for more complex trig expressions
Finding the slope of a trigonometric curve at a specific point

Step-by-Step Example

Problem: Differentiate y = 3sin x − 2cos x + tan x

Differentiate each term separately

Apply the sum rule and each trig derivative in turn.

d/dx(3sin x) = 3cos x

Differentiate the cosine term

The two negatives (from -2 and from cos's own derivative) cancel.

d/dx(-2cos x) = -2·(-sin x) = 2sin x

Differentiate the tangent term

Use the standard tan x derivative.

d/dx(tan x) = sec²x

✓

Answer: dy/dx = 3cos x + 2sin x + sec²x

Interactive Calculator

x (radians)

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Common Mistakes

Mistake: Forgetting the negative sign on d/dx(cos x).
Fix: d/dx(cos x) = −sin x, with a negative sign — a very commonly dropped detail.
Mistake: Mixing up which functions have negative derivatives.
Fix: The three "co-" functions — cos, cot, csc — all have negative derivatives. sin, tan, and sec do not.

Practice Questions

Differentiate y = 4cos x + 5tan x.
What is d/dx(cot x)?

Frequently Asked Questions

How can I prove d/dx(sin x) = cos x?

It's proven from first principles using the limit definition of the derivative combined with the squeeze theorem applied to sin x's limit behavior near 0.

Why do sec x and cot x involve products rather than simple squares?

Because sec x and csc x aren't purely reciprocal-squared like tan x and cot x are in their own derivatives — their derivatives come out as products (sec x·tan x and −csc x·cot x respectively) when derived via the quotient rule.