What Is the Derivative of cos x?
Short answer: −sin x. But knowing the answer and knowing how to use it — and where the negative sign actually comes from — are two different things. This guide walks through the result, the proof, chain rule extensions, and the errors that cost students marks on calculus exams.
The derivative of cos x is −sin x. That’s the answer your textbook gives you, usually in a table, with no explanation. Most students memorise it and move on — until the exam asks them to prove it, or to differentiate something like cos(3x² + 1), and suddenly the memorised result isn’t enough. So here’s what you actually need to know: the result, where it comes from, how to apply it with the chain rule, and the specific mistakes that cause students to lose marks.
What This Guide Covers
The Result — d/dx [cos x] = −sin x
This is the core fact. Memorise it. Then understand it.
What it means in plain language: at any point on the cosine curve, the slope of the tangent line at that point equals the negative of the sine of x at that point. When x = 0, cos x is at its peak and −sin(0) = 0 — so the slope is zero there. That checks out: the cosine curve is flat at the top. When x = π/2, the cosine curve is heading steeply downward — and −sin(π/2) = −1. The slope is −1 at that point. These sanity checks are worth doing because they show you the formula actually makes geometric sense.
The two trig functions are linked through differentiation: d/dx [sin x] = cos x and d/dx [cos x] = −sin x. If you differentiate cosine twice, you get back to negative cosine. Differentiate four times and you’re back to cosine itself. This cycling property is one reason trig functions appear so often in physics and engineering — they naturally describe oscillating systems.
Where the Negative Sign Comes From
The negative sign is not arbitrary. It reflects something real about the cosine function’s behaviour: the cosine is decreasing whenever the sine is positive, and increasing whenever the sine is negative. Slope and sine always have opposite signs for cosine. That relationship is exactly what −sin x captures.
The Geometric View
Sketch the cosine curve from 0 to 2π. At x = 0, it’s flat — slope zero. Moving right, the curve falls. The slope is negative. The sine of x in that interval is positive. So slope = −sin x holds: negative slope, positive sine. After x = π, cosine rises again — positive slope — and sine is now negative. Negative times negative = positive. Still consistent.
The Algebraic View
When you expand cos(x + h) using the angle addition formula and apply the limit definition of the derivative, one of the resulting limit terms evaluates to −sin x — not +sin x. The negative is baked into the angle addition identity for cosine. The algebra forces it.
First Principles Proof, Step by Step
Some courses require you to derive this result from scratch using the limit definition. Here’s how to do it, broken into clear steps you can reproduce on an exam.
The limit definition of a derivative is:
Apply this to f(x) = cos x:
This is just the definition. Substitute f(x) = cos x directly. Nothing happens yet.
This identity must be known — it appears on many formula sheets but professors often require you to apply it without the sheet on in-class tests. Substitute it in: the numerator becomes (cos x · cos h − sin x · sin h − cos x).
Group terms with cos x together and terms with sin x together. Since cos x and sin x don’t depend on h, they can be pulled outside the limits. Now you have two standard trig limits to evaluate.
These two limits are standard results that you either prove separately or cite as known. The first says cosine barely changes near h = 0 — the change is second-order. The second says sine and its argument are nearly equal near zero. Both appear in any calculus textbook and are verified numerically at Khan Academy’s calculus resources.
The first term vanishes. The second term gives −sin x. That’s the result. The proof is complete.
Follow the five steps above exactly: definition → angle addition identity → factor and split → apply the two trig limits → combine. State each trig limit explicitly — don’t just say “this limit equals 1.” Examiners want to see that you know which limits you’re using and why. The whole proof fits in about 8 lines of working.
Khan Academy’s calculus unit on derivatives of trigonometric functions covers both the first-principles derivation of d/dx [sin x] and d/dx [cos x] and the two fundamental trig limits used in those proofs. The two limits — lim(h→0) sin(h)/h = 1 and lim(h→0) (cos h − 1)/h = 0 — are derived geometrically using the squeeze theorem before being applied to the derivative proofs. If your course requires geometric justification for these limits (not just algebraic citation), the Khan Academy treatment is a reliable reference: khanacademy.org/math/ap-calculus-ab/ab-differentiation-1-new/ab-2-7/v/derivatives-of-sinx-and-cosx.
Chain Rule: cos(ax), cos(x²), and More
Knowing d/dx [cos x] = −sin x gets you through basic problems. The chain rule is what you need for everything else — and it’s where most assignment marks are actually won or lost.
If y = cos(u) where u is a function of x, then dy/dx = −sin(u) · du/dx. Differentiate the outside (cosine → negative sine), keep the inside unchanged in the sine, then multiply by the derivative of the inside.
d/dx [cos(ax)] = −a sin(ax)
The inner function is u = ax. Its derivative is a. Apply the chain rule: differentiate the cosine to get −sin(ax), then multiply by a.
Example: d/dx [cos(3x)] = −3 sin(3x)Example: d/dx [cos(−2x)] = 2 sin(−2x) — note: a = −2, so −a = 2
Example: d/dx [cos(x/4)] = −(1/4) sin(x/4)
d/dx [cos(x²)] = −2x sin(x²)
The inner function is u = x². Its derivative is 2x. The outside differentiates to −sin(x²). Multiply together.
Example: d/dx [cos(x³ + 5x)] = −(3x² + 5) sin(x³ + 5x)Example: d/dx [cos(√x)] = d/dx [cos(x^½)] = −(1/(2√x)) sin(√x)
d/dx [cos²x] = −2 cos x · sin x = −sin(2x)
Here cos²x means (cos x)². The outer function is u² and the inner function is cos x. Apply chain rule: 2(cos x) · (−sin x) = −2 cos x sin x. Using the double angle identity, this simplifies to −sin(2x).
Note: cos²x and cos(x²) are completely different expressions. The first is the square of cosine. The second is cosine of the square. Don’t mix them up.d/dx [x · cos x] = cos x − x sin x
Use the product rule: d/dx [uv] = u’v + uv’. Let u = x (so u’ = 1) and v = cos x (so v’ = −sin x). Result: 1 · cos x + x · (−sin x) = cos x − x sin x.
Example: d/dx [x² cos x] = 2x cos x − x² sin xExample: d/dx [eˣ cos x] = eˣ cos x − eˣ sin x = eˣ(cos x − sin x)
| Expression | Derivative | Rule Used |
|---|---|---|
| cos x | −sin x | Standard result |
| cos(5x) | −5 sin(5x) | Chain rule |
| cos(x²) | −2x sin(x²) | Chain rule |
| cos²x | −sin(2x) | Chain rule + double angle identity |
| x cos x | cos x − x sin x | Product rule |
| cos x / x | (−x sin x − cos x) / x² | Quotient rule |
| eˣ cos x | eˣ(cos x − sin x) | Product rule |
| cos(sin x) | −sin(sin x) · cos x | Chain rule (trig inside trig) |
The Second Derivative of cos x
The second derivative means differentiating twice. Start with cos x, differentiate once to get −sin x, then differentiate again.
That’s −cos x. The cosine comes back, but negated. This result matters because it means cosine satisfies the differential equation y” = −y. That equation describes simple harmonic motion — springs, pendulums, electrical oscillators. Every time a physics course brings in cosine to model oscillation, this is exactly why.
The Full Derivative Cycle
- 1st derivative: d/dx [cos x] = −sin x
- 2nd derivative: d²/dx² [cos x] = −cos x
- 3rd derivative: d³/dx³ [cos x] = sin x
- 4th derivative: d⁴/dx⁴ [cos x] = cos x
After four differentiations, you’re back where you started. The cycle repeats every four steps.
Why This Matters for Exams
Second derivative problems often ask you to find concavity or inflection points. For y = cos x, set −cos x = 0 to find inflection points. That gives x = π/2, 3π/2, … — the same x-values where cosine crosses zero. Check concavity by testing sign of −cos x in each interval.
Related Derivatives to Know Alongside cos x
If you know the derivative of cosine, you’re halfway to knowing all six trig derivatives. They’re paired — and once you know the patterns, you don’t need to memorise each one independently.
Paired as: Trig Function → Derivative
sin x → cos x (no negative)cos x → −sin x (negative, switches to sine)
tan x → sec²x
cot x → −csc²x (notice the negative mirrors cos → −sin)
sec x → sec x tan x
csc x → −csc x cot x (again, co-function takes the negative)
Pattern: every “co-” function (cosine, cotangent, cosecant) introduces a negative sign when differentiated. This is a reliable exam shortcut.
Common Exam Mistakes — and How to Avoid Them
Writing +sin x Instead of −sin x
The most common error. Students confuse the derivative of cosine with the derivative of sine. d/dx [sin x] = +cos x, but d/dx [cos x] = −sin x. The negative is not optional.
Remember: Co-Functions Get the Negative
When you differentiate any co-function (cos, cot, csc), a negative sign appears. Build this into your memory. If you write +sin x as the derivative of cos x, flag it immediately as almost certainly wrong.
Forgetting the Chain Rule Multiplier
Writing d/dx [cos(3x)] = −sin(3x) without the coefficient 3. The inner function is 3x; its derivative is 3. The chain rule requires that 3 to appear out front: −3 sin(3x).
Check the Inner Function Every Time
Any time the argument of cosine is not a bare x, you need the chain rule. Ask: what’s inside the cosine? What’s its derivative? Then multiply. This check should be automatic.
Confusing cos²x and cos(x²)
These are different expressions with different derivatives. cos²x = (cos x)², so its derivative is −2 cos x sin x. cos(x²) has inner function x², giving derivative −2x sin(x²). Students often apply the wrong chain rule structure.
Identify the Structure Before Differentiating
Before you differentiate anything, write out: what is the outer function? What is the inner function? For cos²x: outer = u², inner = cos x. For cos(x²): outer = cos(u), inner = x². The structure determines the method.
Differentiating cos x and Getting cos x
This happens when students mix up integration and differentiation. The integral of sin x is −cos x + C. The derivative of cos x is −sin x. They’re related, not the same operation.
Know the Direction: Are You Differentiating or Integrating?
Differentiation of cos x = −sin x. Integration of cos x = sin x + C. They’re inverses of each other. Read the question. The operation sign or notation (d/dx vs ∫) tells you which direction you’re going.
Practice Problems with Guidance
These are the types of problems that appear on calculus assignments and exams. Work through each one before reading the guidance.
Find d/dx [cos(4x − 1)]
What to do: Identify the inner function: u = 4x − 1. Its derivative is 4. Differentiate the outer cosine to get −sin(4x − 1), then multiply by 4. Answer: −4 sin(4x − 1).
Find d/dx [x³ cos x]
What to do: Product rule. Let u = x³ (u’ = 3x²) and v = cos x (v’ = −sin x). Apply u’v + uv’: 3x² cos x + x³(−sin x). Answer: 3x² cos x − x³ sin x.
Find d/dx [cos(e^x)]
What to do: Chain rule with an exponential inner function. The outer is cosine, the inner is eˣ. Derivative of eˣ is eˣ. Answer: −eˣ sin(eˣ). The eˣ comes from the chain rule multiplier.
Find the second derivative of y = cos(2x)
What to do: First derivative of cos(2x) is −2 sin(2x) (chain rule). Now differentiate −2 sin(2x): the derivative of sin(2x) is 2cos(2x), so the second derivative is −2 · 2 cos(2x). Answer: −4 cos(2x).
Prove that y = cos x satisfies y” + y = 0
What to do: Find y” = −cos x (from the second derivative result). Substitute: y” + y = −cos x + cos x = 0. Done. This type of “verify the differential equation” question is common in both calculus and differential equations courses.
The issues are usually one of three things: the chain rule isn’t automatic yet, you’re not identifying inner and outer functions clearly before starting, or you’re mixing up differentiation and integration rules. If it’s any of these, see calculus homework help for structured support, or math assignment help if you need assistance with a specific problem set or exam prep.
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Before Your Next Calculus Problem Set
The derivative of cos x is −sin x. That one fact, applied cleanly through the chain rule and product rule, covers the majority of trig differentiation problems you’ll see in a standard calculus course.
What trips students up isn’t the result itself — it’s the chain rule setup. Before differentiating anything involving cosine, pause and identify what’s inside the cosine and what its derivative is. One second of structure-checking prevents most of the errors above.
If you’ve been assigned a problem set that goes beyond basic derivatives — implicit differentiation involving cosine, differential equations, or Fourier analysis — see math assignment help for support with the written and working components of those assignments.
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