AHL 5.15 (HL)—Further derivatives and integrals

Syllabus
First assessment 2021
Objective
—
Level
HL

Further derivative pairs unlock advanced integrals

HL only

Further derivative pairs unlock advanced integrals.

Key derivatives are (tan⁡x)′=sec⁡2x(\tan x)'=\sec^2x, (sec⁡x)′=sec⁡xtan⁡x(\sec x)'=\sec x\tan x, (cosec⁡x)′=−cosec⁡xcot⁡x(\cosec x)'=-\cosec x\cot x, (cot⁡x)′=−cosec⁡2x(\cot x)'=-\cosec^2x, (ax)′=axln⁡a(a^x)'=a^x\ln a, (log⁡ax)′=1/(xln⁡a)(\log_a x)'=1/(x\ln a), (arcsin⁡x)′=1/1−x2(\arcsin x)'=1/\sqrt{1-x^2}, (arccos⁡x)′=−1/1−x2(\arccos x)'=-1/\sqrt{1-x^2} and (arctan⁡x)′=1/(1+x2)(\arctan x)'=1/(1+x^2).

Example

Because 1/[(x+1)(x+2)]=1/(x+1)−1/(x+2)1/[(x+1)(x+2)]=1/(x+1)-1/(x+2), partial fractions give ∫dx/[(x+1)(x+2)]=ln⁡∣x+1∣−ln⁡∣x+2∣+C\int dx/[(x+1)(x+2)]=\ln|x+1|-\ln|x+2|+C. For ∫dx/[1+(2x+1)2]\int dx/[1+(2x+1)^2], the linear inner derivative gives 12arctan⁡(2x+1)+C\tfrac12\arctan(2x+1)+C.

Match an integrand to a derivative pair, include the reciprocal inner-gradient factor for a linear composite, and use partial fractions before integrating a rational expression when required.

Inverse-trig derivative domains and logarithmic absolute values matter; a memorised form without its domain or inner-gradient factor is incomplete.