5.15 Further derivatives and integrals (IB Maths: Analysis and Approaches HL): Revision notes

Section 1

Derivatives of the reciprocal trigonometric functions

With $\sec x = \dfrac{1}{\cos x}$ , $\operatorname{cosec}x = \dfrac{1}{\sin x}$ , $\cot x = \dfrac{1}{\tan x}$ : $\frac{d}{dx}\tan x = \sec^{2}x,\quad \frac{d}{dx}\sec x = \sec x\tan x,$ $\frac{d}{dx}\operatorname{cosec}x = -\operatorname{cosec}x\cot x,\quad \frac{d}{dx}\cot x = -\operatorname{cosec}^{2}x.$ These are in the formula booklet, and each follows from the quotient or chain rule, e.g. $\dfrac{d}{dx}(\cos x)^{-1} = (\cos x)^{-2}\sin x = \sec x\tan x$ . Combine them with the chain rule: $\dfrac{d}{dx}\operatorname{cosec}3x = -3\operatorname{cosec}3x\cot 3x$ .

Key termssecantcosecantcotangent

Exam tip

The 'co-' functions ( $\operatorname{cosec}$ , $\cot$ ) have derivatives with a minus sign.

Section 2

Exponentials and logarithms with base a

Since $a^{x} = e^{x\ln a}$ , $\frac{d}{dx}a^{x} = a^{x}\ln a, \qquad \frac{d}{dx}\log_{a}x = \frac{1}{x\ln a}.$ For example $\dfrac{d}{dx}3^{x} = 3^{x}\ln 3$ and $\dfrac{d}{dx}\log_{2}(x^{2} + 1) = \dfrac{2x}{(x^{2} + 1)\ln 2}$ .

Reversing: $\displaystyle\int a^{x}\,dx = \dfrac{a^{x}}{\ln a} + C$ .

Key termsbase a

Common mistake

Using the power rule on $3^{x}$ : $x\cdot3^{x-1}$ is wrong because the exponent is the variable.

Section 3

Inverse trigonometric functions

$\frac{d}{dx}\arcsin x = \frac{1}{\sqrt{1 - x^{2}}},\quad \frac{d}{dx}\arccos x = -\frac{1}{\sqrt{1 - x^{2}}},\quad \frac{d}{dx}\arctan x = \frac{1}{1 + x^{2}}.$ With the chain rule: $\dfrac{d}{dx}\arctan(2x) = \dfrac{2}{1 + 4x^{2}}$ , $\dfrac{d}{dx}\arcsin\dfrac{x}{3} = \dfrac{1}{\sqrt{9 - x^{2}}}$ .

The matching integrals (in the booklet) are $\int\frac{dx}{a^{2} + x^{2}} = \frac{1}{a}\arctan\frac{x}{a} + C, \qquad \int\frac{dx}{\sqrt{a^{2} - x^{2}}} = \arcsin\frac{x}{a} + C.$

Key termsarcsinarctan

Common mistake

Confusing $\tan^{-1}x$ (arctan) with $\dfrac{1}{\tan x}$ ( $\cot x$ ).

Section 4

The indefinite integral as a family of curves

Every function in this topic can be integrated in reverse, including composites with a linear function: $\displaystyle\int\sec^{2}(3x)\,dx = \dfrac{1}{3}\tan 3x + C$ , $\displaystyle\int\sec x\tan x\,dx = \sec x + C$ .

Because of the $+\,C$ , an indefinite integral describes a family of curves, each a vertical translation of the others. A boundary condition (a known point) picks out one member. For $\dfrac{dh}{dt} = \dfrac{20}{(t-2)^{2} + 4}$ with $h(0) = 5$ : $h = 10\arctan\dfrac{t - 2}{2} + 5 + \dfrac{5\pi}{2}$ .

Key termsfamily of curvesboundary condition

Section 5

Partial fractions and completing the square

If the denominator factorises, split into partial fractions and integrate to logarithms: $\frac{x + 7}{(x - 1)(x + 3)} = \frac{2}{x - 1} - \frac{1}{x + 3} \Rightarrow \int = 2\ln|x - 1| - \ln|x + 3| + C.$ If a quadratic denominator does not factorise, complete the square to reach an arctan form: $\int\frac{dx}{x^{2} + 2x + 5} = \int\frac{dx}{(x + 1)^{2} + 2^{2}} = \frac{1}{2}\arctan\frac{x + 1}{2} + C.$ Check the discriminant: $b^{2} - 4ac \ge 0$ means factorise, $< 0$ means complete the square.

Key termspartial fractionscompleting the square

Common mistake

Dropping the factor $\dfrac{1}{a}$ in $\dfrac{1}{a}\arctan\dfrac{x}{a}$ .

Exam tip

Combine logarithms at the end with the laws of logarithms to reach a single $\ln k$ .

Must know

$\tan x \to \sec^{2}x$ , $\sec x \to \sec x\tan x$ , $\operatorname{cosec}x \to -\operatorname{cosec}x\cot x$ , $\cot x \to -\operatorname{cosec}^{2}x$ .
$a^{x} \to a^{x}\ln a$ ; $\log_{a}x \to \dfrac{1}{x\ln a}$ .
$\arcsin x \to \dfrac{1}{\sqrt{1 - x^{2}}}$ , $\arccos x \to -\dfrac{1}{\sqrt{1 - x^{2}}}$ , $\arctan x \to \dfrac{1}{1 + x^{2}}$ .
$+\,C$ gives a family of curves; a boundary condition fixes one.
Factorising denominator: partial fractions. Non-factorising: complete the square, then arctan.

That's the notes covered.

Carry on to the next subtopic.