Derivatives of Trigonometric Functions

Push a child on a swing and their height rises and falls like a sine wave; ask how fast they are moving at any instant and you are asking for its derivative. The speed of anything that swings, vibrates or wobbles is exactly what the trig derivatives on this page deliver.

Trace the graph of \sin x with your finger. It climbs, eases off, curls over at the top, tips back down, bottoms out, and climbs again — a smooth, endless swell. Now ask the calculus question: how steep is it at each point? Plot that slope, moment by moment, and something almost magical happens — the slope graph is another wave of the exact same shape, just slid over a quarter turn. It's \cos x.

Differentiate \sin x and you get \cos x; differentiate \cos x and you get -\sin x:

\frac{d}{dx}\sin x = \cos x, \qquad \frac{d}{dx}\cos x = -\sin x

Differentiating doesn't leave the trig family — it just shifts you around it. That is not a coincidence; it is the deep reason sine waves describe things that oscillate. A quantity whose rate of change keeps feeding back into itself like this can never settle down — it has to keep swinging. That is the mathematics of a plucked string, a swinging pendulum, a note of music, and the light reaching your eye right now.

The picture behind it: slope becomes height

The slope of \sin x at each point is exactly the height of \cos x there. Check it at three places:

That is why differentiating slides the wave a quarter period to the left: the cosine curve is just the sine curve reporting on its own steepness.

Tangent, too

The tangent function fits the same family, with a slightly busier derivative:

\frac{d}{dx}\tan x = \sec^2 x

You can see where this comes from: \tan x = \tfrac{\sin x}{\cos x}, and applying the quotient rule to that fraction tidies up (using \sin^2 x + \cos^2 x = 1) into the neat \dfrac{1}{\cos^2 x} = \sec^2 x. Because \sec^2 x \ge 1 everywhere it is defined, the tangent curve is always climbing — it never turns over.

For the three standard trig functions:

See the slope become the height

Here is y = \sin x (solid) drawn over a few periods, with its derivative y' = \cos x (dashed). Pick any point on the solid wave: how steeply it climbs there is exactly the height of the dashed wave at the same x. Where \sin x is flattest at its peaks, \cos x crosses zero; where \sin x rises fastest through the origin, \cos x is at its highest.

Worked example 1 — the chain rule on \sin(3x)

A great deal of real trig calculus is \sin(kx), not bare \sin x — the k sets how fast the wave wobbles. Differentiate y = \sin(3x).

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

The 3 pops out the front. The same move gives \dfrac{d}{dx}\cos(5x) = -5\sin(5x) — outer derivative -\sin, times the inner derivative 5.

Worked example 2 — a product, x\cos x

Differentiate y = x\cos x. This is one thing times another, so reach for the product rule, (uv)' = u'v + uv', with u = x and v = \cos x.

\frac{d}{dx}\big(x\cos x\big) = (1)(\cos x) + (x)(-\sin x) = \cos x - x\sin x.

Notice how the -\sin x from differentiating cosine carries its minus sign right through to the answer. Lose that sign and the whole thing is wrong.

Worked example 3 — finding a stationary point

Where does y = \sin x have a stationary point (a peak or trough) between 0 and 2\pi? A stationary point is where the slope is zero, so set the derivative to zero:

y' = \cos x = 0 \;\Rightarrow\; x = \frac{\pi}{2}\ \text{ or }\ x = \frac{3\pi}{2}.

At x = \tfrac{\pi}{2}, \sin x = 1 — the maximum. At x = \tfrac{3\pi}{2}, \sin x = -1 — the minimum. The derivative found the crest and the trough of the wave for us, exactly where the tangent line goes flat. This is the everyday job of a trig derivative: locating the high and low points of anything that oscillates.

Every rule on this page — \tfrac{d}{dx}\sin x = \cos x and its siblings — is only correct when x is in radians. In degrees they are simply wrong. Differentiate \sin(x^\circ) and an ugly conversion factor crawls out:

\frac{d}{dx}\sin(x^\circ) = \frac{\pi}{180}\cos(x^\circ).

That stray \tfrac{\pi}{180} \approx 0.01745 is the fingerprint of the wrong unit. It appears because the clean limit \displaystyle\lim_{h\to 0}\tfrac{\sin h}{h} = 1 — the fact the whole derivative rests on — is only 1 when h is in radians. This is the real reason mathematicians call radians the natural unit for angles: they are the choice that makes calculus come out clean. So set your calculator to radians before you differentiate.

And two more classic slips: don't lose the minus on \tfrac{d}{dx}\cos x = -\sin x, and don't forget the chain rule — \tfrac{d}{dx}\sin(kx) = k\cos(kx) has a factor of k out front, not just \cos(kx).

Keep differentiating and watch the four-step loop:

\sin x \;\to\; \cos x \;\to\; -\sin x \;\to\; -\cos x \;\to\; \sin x \;\to\;\cdots

Four turns of the crank and you are back where you started. That closed loop is the mathematical signature of oscillation. Look at the second derivative:

\frac{d^2}{dx^2}\sin x = -\sin x.

Read that as a sentence: "the acceleration of this quantity is the negative of the quantity itself." Whenever something gets pulled back towards the middle in proportion to how far it has strayed — a mass on a spring, a pendulum, the air pressure in a sound wave, the current in an AC circuit, the field of a light wave, even a quantum wavefunction — it obeys exactly this equation, and the answer must be a sine or cosine. A system whose rate of change feeds back on itself this way has no choice but to oscillate. That single feedback loop, captured by these little derivative rules, is the heart of every equation for waves and vibrations in physics.

See it explained