Differentiating Reciprocal Trigonometric Functions (DP IB Analysis & Approaches (AA)): Revision Note

Written by: Paul

Reviewed by: Dan Finlay

Updated on 15 July 2025

Differentiating reciprocal trigonometric functions

What are the reciprocal trigonometric functions?

Secant, cosecant and cotangent are abbreviated and defined as

$\sec x = \frac{1}{\cos x}$ $cosec x = \frac{1}{\sin x}$ $\cot x = \frac{1}{\tan x}$

Remember that for calculus, angles need to be measured in radians
- $θ$ may be used instead of $x$
$cosec x$ is sometimes further abbreviated to $\csc x$

What are the derivatives of the reciprocal trigonometric functions?

$f (x) = \sec x$
- $f^{'} (x) = \sec x \tan x$
$f (x) = cosec x$
- $f^{'} (x) = - cosec x \cot x$
$f (x) = \cot x$
- $f^{'} (x) = - {cosec}^{2} x$

Examiner Tips and Tricks

These three derivatives are given in the exam formula booklet.

How do I show or prove the derivatives of the reciprocal trigonometric functions?

For $y = \sec x$
- Rewrite, $y = \frac{1}{\cos x}$
- Use quotient rule, $\frac{d y}{d x} = \frac{\cos x (0) - (1) (- \sin x)}{\cos^{2} x}$
- Rearrange, $\frac{d y}{d x} = \frac{\sin x}{\cos^{2} x}$
- Separate, $\frac{d y}{d x} = \frac{1}{\cos x} \times \frac{\sin x}{\cos x}$
- Rewrite, $\frac{d y}{d x} = \sec x \tan x$
Similarly, for $y = cosec x$
- $y = \frac{1}{\sin x}$
- $\frac{d y}{d x} = \frac{\sin x (0) - (1) \cos x}{\sin^{2} x}$
- $\frac{d y}{d x} = \frac{- \cos x}{\sin^{2} x}$
- $\frac{d y}{d x} = - \frac{1}{\sin x} \times \frac{\cos x}{\sin x}$
- $\frac{d y}{d x} = - cosec x \cot x$
For the derivative of $y = \cot x$ , see the Worked Example

What do the derivatives of reciprocal trig functions look like with linear functions of x?

For linear functions of the form ax+b
- $f (x) = \sec (a x + b)$
  - $f^{'} (x) = a \sec (a x + b) \tan (a x + b)$
- $f (x) = cosec (a x + b)$
  - $f^{'} (x) = - a cosec (a x + b) \cot (a x + b)$
- $f (x) = \cot (a x + b)$
  - $f^{'} (x) = - a {cosec}^{2} (a x + b)$
- These are not given in the exam formula booklet
  - they can be derived from chain rule
  - they are not essential to remember

Examiner Tips and Tricks

Even if you think you have remembered these derivatives, always use the formula booklet to double check. The squares and negatives are easy to get muddled up!

Where two trig functions are involved in the derivative be careful with angle multiples like $x, 2 x, 3 x$ , etc. An example of a common mistake is differentiating $y = cosec 3 x$ , and getting $\frac{d y}{d x} = - 3 cosec x \cot 3 x$ instead of $\frac{d y}{d x} = - 3 cosec 3 x \cot 3 x$ .

Worked Example

Curve C has equation $y = 2 \cot (3 x - \frac{π}{8})$ .

a) Show that the derivative of $\cot x$ is $- {cosec}^{2} x$ .

b) Find $\frac{d y}{d x}$ for curve C.

c) Find the gradient of curve C at the point where $x = \frac{7 π}{24}$ .

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Differentiating Reciprocal Trigonometric Functions (DP IB Analysis & Approaches (AA)): Revision Note

Differentiating reciprocal trigonometric functions

What are the reciprocal trigonometric functions?

What are the derivatives of the reciprocal trigonometric functions?

How do I show or prove the derivatives of the reciprocal trigonometric functions?

What do the derivatives of reciprocal trig functions look like with linear functions of x?

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