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Differentiating Trigonometric Functions

Combining the quotient and chain rules gives the derivatives of tan, sec, cosec and cot. These standard results should be memorised for exam speed.

26 min Differentiation

Edexcel A level Maths: 9.6 Differentiating Trigonometric Functions (Sec, Cosec, Cot and Tan)Curated by Kepler, not produced by us. Watch it before the notes below.

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What you'll be able to do

Know derivatives of tan, sec, cosec, cot
Derive them from sin and cos
Apply with the chain rule
Use them in larger problems

The standard derivatives

These follow from the quotient rule applied to the reciprocal functions:

\frac{d}{d x} (tan x) = sec^{2} x

Tangent.

\frac{d}{d x} (sec x) = sec x tan x

Secant.

\frac{d}{d x} (cosec x) = - cosec x cot x

Cosecant.

\frac{d}{d x} (cot x) = - cosec^{2} x

Cotangent.

With the chain rule

For a function inside, multiply by its derivative, e.g. $\frac{d}{d x} tan (3 x) = 3 sec^{2} (3 x)$ .

1Outer:

sec u \to sec u tan u

. Inner:

u = 2 x

u^{'} = 2

\frac{d y}{d x} = 2 sec 2 x tan 2 x

Answer

2 sec 2 x tan 2 x

Tip — The “co” functions (cosec, cot) carry a minus sign.

Formula recap

\frac{d}{d x} (tan x) = sec^{2} x

tan.

\frac{d}{d x} (sec x) = sec x tan x

sec.

\frac{d}{d x} (cot x) = - cosec^{2} x

cot (minus).

Common mistakes to avoid

Giving d/dx(cot x) = cosec²x (no minus).

It is −cosec²x.

Forgetting the chain rule on tan(kx) etc.

Multiply by the derivative of the inside.

Key takeaways

tan→sec²x, sec→sec x tan x.
cosec→−cosec x cot x, cot→−cosec²x.
Chain rule: multiply by the inner derivative.

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Differentiating Trigonometric Functions

The standard derivatives

With the chain rule

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Differentiating Trigonometric Functions

The standard derivatives

With the chain rule

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