Teacher page: further derivatives and partial fractions (AA 5.15)

The one thing to do with the figure

Let them verify the table rather than trust it.

Pick arcsin and move x towards 1: the gradient climbs towards infinity, which is exactly what 1/√(1 − x²) says and what the graph of arcsin looks like.

A table that has been checked is remembered differently from a table that was handed over.

Three routes, and the choice is the skill

A fraction with a quadratic denominator needs one of: the reverse chain rule (if the top is the derivative of the bottom), partial fractions (if the bottom factorises), or completing the square into an arctan (if it does not).

Deciding which takes ten seconds and determines the whole question. Spend the lesson on the decision, not the algebra.

The answers

Key derivatives	tan → sec²; arctan → 1/(1+x²); arcsin → 1/√(1−x²); a^x → a^xln a.
Completing the square	∫1/(x²+2x+5) dx = ½arctan((x+1)/2) + C.
Partial fractions	1/((x+1)(x+2)) = 1/(x+1) − 1/(x+2), integrating to ln\|(x+1)/(x+2)\| + C.
1, 2, 3	0.5; B, factorise and split; B, ½tan(2x+5) + C.

Where the marks go

1 markChoosing the right route for the fraction.

1 markSetting up and solving the partial fractions correctly.

1 markThe dividing constant on a composite with a linear function.

1 markThe modulus inside every logarithm.

What each wrong answer tells you

They give	What it means
Completing the square on x² − 4	It factorises, so partial fractions is the route. Completing the square leads nowhere useful.
arctan with a root in it	Confusing the arctan and arcsin derivatives. The root belongs to arcsin.
Missing minus on arccos	Every co-function picks up a minus. Worth stating as a pattern rather than four separate facts.
Dividing by the wrong constant	Composite with 2x + 5 divides by 2, not 5.

Other things they will say

"Do I have to memorise all of these?" The inverse trigonometric ones and tan, yes. The others follow from the chain rule or appear in the formula booklet; check which for your session.

"When do I complete the square?" When the quadratic has no real factors. Check the discriminant first; it is faster than trying.

"Why does arcsin blow up at 1?" Because the graph of arcsin is vertical there. The formula and the picture agree, which is a good moment to say so.

A possible order

	What is happening
1	The table, checked with the widget rather than recited.
2	Reading it backwards as integrals, with linear composites.
3	Completing the square into an arctan.
4	Partial fractions, with the factorise-first decision.
5	Questions 1 to 3.
6	A mixed set where they must choose the route.

Two things not to say

Do not teach the three routes in three separate lessons. The skill is telling them apart, and that only exists when they are side by side.

Do not let them skip factorising the denominator before choosing. It is the ten seconds that decides everything after it.

Questions to set

Three tiers, ramping the way practice should: the method on its own, then the method inside something real, then a challenge. Set the tier the class in front of you needs rather than one undifferentiated sheet. Answers are given so these can go straight onto a board.

1Fluency

The method on its own, with friendly numbers. Set these first and move on quickly once they are secure.

Trig derivativeDifferentiate tan x and evaluate at x = 0.
sec²x, which is 1 at x = 0.
An inverseDifferentiate arctan x and evaluate at x = 1.
1/(1 + x²), which is 0.5 at x = 1.

2In context

The same skill inside a real situation, where the first job is working out what is being asked.

Another inverseDifferentiate arcsin x and evaluate at x = 0.5.
1/√(1 − x²) = 1/√0.75 = 1.155.
A reciprocal trig functionDifferentiate sec x and evaluate at x = 0.
sec x tan x, which is 0 at x = 0, since tan 0 = 0.

3Challenge

Reasoning, working backwards, or spotting an error. These are where the top grades are decided.

Derive oneObtain the derivative of arcsin x by differentiating sin y = x implicitly.
cos y (dy/dx) = 1, so dy/dx = 1/cos y. With sin y = x, cos y = √(1 − x²), giving 1/√(1 − x²). The positive root is correct because arcsin has range −π/2 to π/2, where cosine is non-negative.
Notice the domainState where the derivative of arcsin x fails to exist and what that means on the graph.
At x = ±1 the denominator is zero, so the derivative is undefined. The graph of arcsin has vertical tangents at those endpoints, which is why the derivative blows up rather than being merely large.

Practicalities

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