The six trigonometric functions differentiate into each other in a short, closed cycle.
\[\frac{d}{dx}\sin x = \cos x\]
What it says
Everything follows from one limit, \(\lim_{h \to 0}\frac{\sin h}{h} = 1\), which says that near zero the sine of an angle is the angle itself. Feeding that into the difference quotient for \(\sin x\) produces \(\cos x\), and the rest of the family follows by the quotient rule.
The full table: \(\left(\sin x\right)' = \cos x\), \(\left(\cos x\right)' = -\sin x\), \(\left(\tan x\right)' = \sec^{2}x\), \(\left(\sec x\right)' = \sec x\tan x\), \(\left(\csc x\right)' = -\csc x\cot x\), \(\left(\cot x\right)' = -\csc^{2}x\). Every co-function carries a minus sign — a reliable memory aid.
Differentiating \(\sin\) four times returns to \(\sin\), which is why higher derivatives of trigonometric functions cycle with period four.
When it applies
Any of the six trigonometric functions applied to the variable.
With the chain rule when the angle is a compound expression, as in \(\sin(3x)\).
All of it assumes radians — in degrees an extra factor of \(\frac{\pi}{180}\) appears.
Five worked examples
Every line is the step the calculator would show, in the order it applies them. Each graph
is live: hover it to read both curves and see the tangent whose slope is the derivative,
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1
Example 1
\[\frac{d}{dx}\left[\sin\left(x\right)\right]\]
Straight from the table.
\[\frac{d}{dx}\left[\sin x\right] = \cos x\]
Answer
\[\cos\left(x\right)\]
The pair runs a quarter-turn apart. f′ peaks where f crosses the axis, the steepest part of the wave, and reads zero at the peaks of f. Drag the crosshair along and watch the two readouts trade places every quarter period.
The sine curve is steepest where the cosine peaks, at \(x = 0\), and flat at its own peaks.
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