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9.6PureStretch

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
1

The standard derivatives

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

ddx(tan⁡x)=sec⁡2x\frac{d}{dx}(\tan x) = \sec^2 x
Tangent.
ddx(sec⁡x)=sec⁡xtan⁡x\frac{d}{dx}(\sec x) = \sec x\tan x
Secant.
ddx(cosec⁡x)=−cosec⁡xcot⁡x\frac{d}{dx}(\operatorname{cosec} x) = -\operatorname{cosec} x\cot x
Cosecant.
ddx(cot⁡x)=−cosec⁡2x\frac{d}{dx}(\cot x) = -\operatorname{cosec}^2 x
Cotangent.
2

With the chain rule

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

Example 1Differentiate y=sec⁡2xy = \sec 2x.
1Outer: sec⁡u→sec⁡utan⁡u\sec u \to \sec u\tan u. Inner: u=2xu = 2x, u′=2u' = 2.
2dydx=2sec⁡2xtan⁡2x\tfrac{dy}{dx} = 2\sec 2x\tan 2x.
Answer
2sec⁡2xtan⁡2x2\sec 2x\tan 2x

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

Formula recap

ddx(tan⁡x)=sec⁡2x\frac{d}{dx}(\tan x) = \sec^2 x
tan.
ddx(sec⁡x)=sec⁡xtan⁡x\frac{d}{dx}(\sec x) = \sec x\tan x
sec.
ddx(cot⁡x)=−cosec⁡2x\frac{d}{dx}(\cot x) = -\operatorname{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.

Test yourself

Answer a few questions, then review anything you missed.

Prefer paper? Open the written self-check

Answer each question before opening its answer. This self-check does not record a quiz score.

  1. ddx(tan⁡x)\frac{d}{dx}(\tan x) equals:

    • sec⁡2x\sec^2 x
    • sec⁡xtan⁡x\sec x\tan x
    • −cosec⁡2x-\operatorname{cosec}^2 x
    • cos⁡x\cos x
    Show answer and explanation

    Answer: sec⁡2x\sec^2 x

    Standard.

  2. ddx(sec⁡x)\frac{d}{dx}(\sec x) equals:

    • sec⁡xtan⁡x\sec x\tan x
    • sec⁡2x\sec^2 x
    • −sec⁡xtan⁡x-\sec x\tan x
    • tan⁡x\tan x
    Show answer and explanation

    Answer: sec⁡xtan⁡x\sec x\tan x

    Standard.

  3. ddx(cot⁡x)\frac{d}{dx}(\cot x) equals:

    • −cosec⁡2x-\operatorname{cosec}^2 x
    • cosec⁡2x\operatorname{cosec}^2 x
    • sec⁡2x\sec^2 x
    • −sec⁡2x-\sec^2 x
    Show answer and explanation

    Answer: −cosec⁡2x-\operatorname{cosec}^2 x

    Note the minus.

  4. ddxtan⁡(3x)\frac{d}{dx}\tan(3x) equals:

    • 3sec⁡2(3x)3\sec^2(3x)
    • sec⁡2(3x)\sec^2(3x)
    • 3sec⁡2x3\sec^2 x
    • sec⁡2(3x)tan⁡(3x)\sec^2(3x)\tan(3x)
    Show answer and explanation

    Answer: 3sec⁡2(3x)3\sec^2(3x)

    Chain rule, ×3.

  5. Which derivatives carry a minus sign?

    • cosec and cot
    • tan and sec
    • all four
    • none
    Show answer and explanation

    Answer: cosec and cot

    The “co” functions.

Next step

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