Derivatives of Trigonometric Functions

The two basic waves have derivatives that loop into each other. 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

There's a picture behind this. The slope of \sin x at each point is exactly the height of \cos x there — which is why differentiating just shifts the wave a quarter turn to the left.

Tangent, too

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

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

For the three standard trig functions:

\dfrac{d}{dx}\sin x = \cos x;
\dfrac{d}{dx}\cos x = -\sin x;
\dfrac{d}{dx}\tan x = \sec^2 x;
these hold only when x is measured in radians;
combine with the chain rule: \dfrac{d}{dx}\sin(kx) = k\cos(kx).

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.