§ 01Prerequisites — Two Essential Limits
Two special trigonometric limits underpin the proofs of d(sin x)/dx and d(cos x)/dx. Everything else follows from these by algebra and the differentiation rules you already know.
Both limits must be evaluated with x measured in radians. The elegant derivative formulas d(sin x)/dx = cos x only hold in radians; in degrees a clumsy factor of π/180 appears everywhere. All work on this page assumes radians.
limh→0 (cos h − 1)/h = 0
Geometric Proof of lim (sin h)/h = 1
Consider a unit circle. For a small angle h (radians), the arc length is h, the chord length is sin h, and the tangent length is tan h. Comparing areas gives the squeeze inequality:
As h → 0: cos h → 1 and 1/cos h → 1
∴ (sin h)/h → 1 by the Squeeze Theorem
Deriving lim (cos h − 1)/h = 0
= (cos²h − 1) / (h(cos h + 1))
= −sin²h / (h(cos h + 1))
= −(sin h)/h × sin h/(cos h + 1)
→ −1 × 0/2 = 0 as h → 0
§ 02Proof from First Principles — d(sin x)/dx
Armed with the two limits, we can differentiate sin x from the definition of the derivative. This is the most important proof in trig differentiation.
Apply the addition formula sin(x+h) = sin x cos h + cos x sin h:
[sin x cos h + cos x sin h − sin x] / hApply cos(x+h) = cos x cos h − sin x sin h:
[cos x cos h − sin x sin h − cos x] / h§ 03All Six Standard Trig Derivatives
With sin and cos established, all four remaining derivatives follow from the quotient rule. Every result is worth memorising, they appear constantly in calculus, physics, and engineering.
| Function f(x) | Derivative f′(x) | How derived |
|---|---|---|
| sin x | cos x | First principles (§ 02) |
| cos x | −sin x | First principles (§ 02) |
| tan x | sec² x | Quotient rule on sin x / cos x |
| cosec x | −cosec x cot x | Quotient rule on 1/sin x |
| sec x | sec x tan x | Quotient rule on 1/cos x |
| cot x | −cosec² x | Quotient rule on cos x / sin x |
d/dx[u/v] = (v u′ − u v′) / v² with u = sin x, v = cos x:
[cos x × cos x − sin x × (−sin x)] / cos² x§ 04Chain Rule Compositions
The most important extension is applying the chain rule when a trig function has a function of x (rather than just x) as its argument. The pattern is: differentiate the outside trig function, multiply by the derivative of the inside.
d/dx [cos(g(x))] = −sin(g(x)) × g′(x)
d/dx [tan(g(x))] = sec²(g(x)) × g′(x)
Outer function: sin( ). Inner function: g(x) = 3x², so g′(x) = 6x.
Write cos&sup5;(x) = [cos(x)]&sup5; and apply the chain rule: outer = u&sup5;, inner = cos(x).
u = x², u′ = 2x v = sin x, v′ = cos x
u = sin x, u′ = cos x v = 1 + cos x, v′ = −sin x
§ 05Inverse Trigonometric Derivatives
The inverse trig functions arcsin, arccos, and arctan are essential in integration (they appear as antiderivatives of algebraic functions). Their derivatives are derived using implicit differentiation on the original trig relationship.
Deriving d(arcsin x)/dx
Differentiate both sides: cos y · dy/dx = 1
dy/dx = 1/cos y = 1/√(1−sin²y) = 1/√(1−x²)
| Function f(x) | Derivative f′(x) | Domain |
|---|---|---|
| arcsin x | 1/√(1−x²) | |x| < 1 |
| arccos x | −1/√(1−x²) | |x| < 1 |
| arctan x | 1/(1+x²) | all x ∈ ℝ |
| arccosec x | −1/(|x|√(x²−1)) | |x| > 1 |
| arcsec x | 1/(|x|√(x²−1)) | |x| > 1 |
| arccot x | −1/(1+x²) | all x ∈ ℝ |
Differentiate f(x) = x arcsin(x) + √(1−x²). This function appears as an antiderivative in integration.
Product rule: u = x, u′ = 1 v = arcsin x, v′ = 1/√(1−x²)
= arcsin x + x/√(1−x²)Chain rule on (1−x²)1/2:
= ½(1−x²)−1/2 × (−2x) = −x/√(1−x²)§ 06Common Mistakes
§ 07Quick Reference & Related Pages
d(cos x)/dx = −sin x
d(tan x)/dx = sec² x
d(cosec x)/dx = −cosec x cot x
d(sec x)/dx = sec x tan x
d(cot x)/dx = −cosec² x
Where Trig Derivatives Lead
§ 08Quiz — Test Your Understanding
Ten questions on trig derivatives, chain rule compositions, and inverse trig. Select the best answer.