Calculus — Derivatives6 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. Each of these can be derived from the limit definition of the derivative and the two functions' relationship through basic trig identities, but in practice they're memorized as a set, since they get used constantly once trig functions start appearing inside more complex expressions.

A useful pattern: sine and cosine are each other's derivatives (with a sign flip going from cosine to sine, but not the other way — differentiating sin x gives cos x, and differentiating that gives −sin x, completing a four-step cycle if you keep going), and every "co-" function (cos, cot, csc) has a negative derivative, while sin, tan, and sec do not.

These six derivatives rarely appear in isolation on a real exam problem — they're almost always combined with the chain rule (when the trig function's argument is itself a function of x, like sin(3x²)) or the product/quotient rule (when a trig function is multiplied or divided with another function). Getting the six base derivatives memorized cold is what makes those more complex combinations manageable.

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
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Step-by-Step Example

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

1
Differentiate each term separately

Apply the sum rule and each trig derivative in turn.

d/dx(3sin x) = 3cos x
2
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
3
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

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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

  1. Differentiate y = 4cos x + 5tan x.

  2. 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.