Introduction to AHL 5.15: Further Derivatives and Integration

In this subtopic, we extend our differentiation and integration toolkit well beyond the basics. You already know how to differentiate polynomials, $sin x$ , $cos x$ , $e^{x}$ , and $ln x$ . Now we tackle:

The remaining trigonometric functions: $tan x$ , $sec x$ , $csc x$ , $cot x$
Exponential and logarithmic functions with arbitrary bases (not just base $e$ )
The three inverse trigonometric functions: $arcsin x$ , $arccos x$ , $arctan x$
Their corresponding indefinite integrals
Partial fraction decomposition as a powerful integration technique

These tools unlock a huge range of functions that appear regularly in IB HL exam questions. Mastery here connects directly to topics in differential equations, kinematics, and mathematical modelling.

Note

All standard derivatives and integrals in this subtopic are provided in the IB formula booklet. However, you must know how to apply them fluently, especially when combined with the chain rule or substitution.

Derivatives of Trigonometric Functions

Beyond $sin x$ and $cos x$ , we have four more trigonometric derivatives to master:

$\frac{d}{d x} (tan x) = sec^{2} x$
$\frac{d}{d x} (sec x) = sec x tan x$
$\frac{d}{d x} (csc x) = - csc x cot x$
$\frac{d}{d x} (cot x) = - csc^{2} x$

Exam Tip

A memory aid for the signs: the derivatives of the co-functions ( $cos x$ , $csc x$ , $cot x$ ) all carry a negative sign. The non-co-functions ( $sin x$ , $sec x$ , $tan x$ ) have positive derivatives.

Example

Example: Differentiating $y = tan (3 x)$ using the chain rule

Let the outer function be $tan (u)$ where $u = 3 x$ .

$\frac{d y}{d x} = sec^{2} (u) \cdot \frac{d u}{d x} = sec^{2} (3 x) \cdot 3 = 3 sec^{2} (3 x)$

Example

Example: Differentiating $y = sec^{2} (x) + cot (5 x)$

Differentiate term by term:

For $sec^{2} (x)$ : treat as $[sec (x)]^{2}$ , use chain rule:
$\frac{d}{d x} [sec^{2} x] = 2 sec x \cdot sec x tan x = 2 sec^{2} x tan x$
For $cot (5 x)$ : use chain rule with $u = 5 x$ :
$\frac{d}{d x} [cot (5 x)] = - csc^{2} (5 x) \cdot 5 = - 5 csc^{2} (5 x)$

So: $\frac{d y}{d x} = 2 sec^{2} x tan x - 5 csc^{2} (5 x)$

Warning

Never forget the chain rule factor! $\frac{d}{d x} (tan (3 x)) = 3 sec^{2} (3 x)$ , not just $sec^{2} (3 x)$ . This is one of the most common errors in HL calculus questions.

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AHL 5.15—Further derivatives and indefinite integration of these, partial fractions

Introduction to AHL 5.15: Further Derivatives and Integration

Derivatives of Trigonometric Functions

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