Paper 2 · Further Pure 2 ✦

Differentiation Extended

New functions to differentiate, second derivatives when $y$ is only defined implicitly or through a parameter, and Maclaurin's series — a polynomial that imitates any function near the origin.

Syllabus 2.3Derivative table given by MF19No general term needed

What the examiner expects you to do

Differentiate hyperbolic functions, and $\sin^{- 1} x, \cos^{- 1} x, \sinh^{- 1} x, \cosh^{- 1} x, \tanh^{- 1} x$ .
Obtain an expression for $\frac{d^{2} y}{d x^{2}}$ when $x$ and $y$ are related implicitly or parametrically.
Derive and use the first few terms of a Maclaurin's series. Deriving a general term is not required, but successive implicit differentiation steps may be.

§1 Differentiating hyperbolic functions

◆ The three on the formula sheet MF19

\frac{d}{d x} \sinh x = \cosh x, \frac{d}{d x} \cosh x = \sinh x, \frac{d}{d x} \tanh x = {sech}^{2} x .

Note the missing minus. $\frac{d}{d x} \cos x = - \sin x$ , but $\frac{d}{d x} \cosh x = + \sinh x$ . Differentiating twice returns you to where you started, rather than flipping the sign — which is exactly why $y = \cosh x$ satisfies $y^{''} = y$ while $y = \cos x$ satisfies $y^{''} = - y$ .

Where they come from (one line each)

\frac{d}{d x} (\frac{e^{x} - e^{- x}}{2}) = \frac{e^{x} + e^{- x}}{2} = \cosh x, \frac{d}{d x} (\frac{e^{x} + e^{- x}}{2}) = \frac{e^{x} - e^{- x}}{2} = \sinh x .

The minus sign from $e^{- x}$ does the work in both. For $\tanh$ , the quotient rule gives

\frac{d}{d x} \frac{\sinh x}{\cosh x} = \frac{\cosh^{2} x - \sinh^{2} x}{\cosh^{2} x} = \frac{1}{\cosh^{2} x} = {sech}^{2} x,

using $\cosh^{2} x - \sinh^{2} x \equiv 1$ . $■$

◇ The reciprocal three not on MF19

\frac{d}{d x} sech x = - sech x \tanh x, \frac{d}{d x} cosech x = - cosech x \coth x, \frac{d}{d x} \coth x = - {cosech}^{2} x .

Each is one chain rule away — for instance $sech x = (\cosh x)^{- 1}$ , so its derivative is $- (\cosh x)^{- 2} \sinh x = - sech x \tanh x$ . Rebuild them rather than memorising, and note that all three carry a minus sign.

§2 Inverse trigonometric and inverse hyperbolic derivatives

◆ The table all on MF19

$f (x)$	$f^{'} (x)$	$f (x)$	$f^{'} (x)$
$\sin^{- 1} x$	$\frac{1}{\sqrt{1 - x^{2}}}$	$\sinh^{- 1} x$	$\frac{1}{\sqrt{1 + x^{2}}}$
$\cos^{- 1} x$	$- \frac{1}{\sqrt{1 - x^{2}}}$	$\cosh^{- 1} x$	$\frac{1}{\sqrt{x^{2} - 1}}$
$\tan^{- 1} x$	$\frac{1}{1 + x^{2}}$	$\tanh^{- 1} x$	$\frac{1}{1 - x^{2}}$

Read the pattern: the trigonometric inverses give $1 - x^{2}$ and $1 + x^{2}$ ; the hyperbolic ones swap those signs over, giving $1 + x^{2}$ , $x^{2} - 1$ and $1 - x^{2}$ .

Deriving them — one method, five functions

Every one uses the same three steps: invert, differentiate implicitly, convert back to $x$ .

$y = \sin^{- 1} x$ : then $x = \sin y$ , so $1 = \cos y \frac{d y}{d x}$ and

\frac{d y}{d x} = \frac{1}{\cos y} = \frac{1}{\sqrt{1 - \sin^{2} y}} = \frac{1}{\sqrt{1 - x^{2}}} .

The positive square root is correct because the principal range is $- \frac{π}{2} \leq y \leq \frac{π}{2}$ , where $\cos y \geq 0$ .

$y = \sinh^{- 1} x$ : then $x = \sinh y$ , so $1 = \cosh y y^{'}$ and $y^{'} = \frac{1}{\cosh y} = \frac{1}{\sqrt{1 + \sinh^{2} y}} = \frac{1}{\sqrt{1 + x^{2}}}$ , using $\cosh^{2} - \sinh^{2} = 1$ . No sign worry here: $\cosh y > 0$ for every $y$ .

$y = \cosh^{- 1} x$ : $x = \cosh y$ gives $y^{'} = \frac{1}{\sinh y} = \frac{1}{\sqrt{\cosh^{2} y - 1}} = \frac{1}{\sqrt{x^{2} - 1}}$ , positive because the principal branch has $y \geq 0$ .

$y = \tanh^{- 1} x$ : $x = \tanh y$ gives $1 = {sech}^{2} y y^{'}$ , so $y^{'} = \frac{1}{{sech}^{2} y} = \frac{1}{1 - \tanh^{2} y} = \frac{1}{1 - x^{2}}$ . $■$

★ A* tip — the chain rule versions you will actually meet

Questions rarely give you a bare $x$ . With the chain rule,

\frac{d}{d x} \sin^{- 1} \frac{x}{a} = \frac{1}{\sqrt{a^{2} - x^{2}}}, \frac{d}{d x} \sinh^{- 1} \frac{x}{a} = \frac{1}{\sqrt{a^{2} + x^{2}}}, \frac{d}{d x} \tan^{- 1} \frac{x}{a} = \frac{a}{a^{2} + x^{2}} .

Each drops out of $\frac{1}{a}$ times the basic derivative evaluated at $x / a$ . These are exactly the shapes that reappear as standard integrals in Topic 11, run backwards.

§3 The second derivative when $y$ is implicit

◆ Two routes — the second is usually cleaner

Differentiate, rearrange, differentiate again. Get $\frac{d y}{d x}$ explicitly, then differentiate that expression with the quotient rule, substituting $\frac{d y}{d x}$ back in wherever it appears.
Differentiate the whole equation twice. Do not rearrange at all — just differentiate the first-derivative equation again, then substitute numbers at the end. This avoids the quotient rule entirely and is far less error-prone.

✎ Worked example — route 1 on a circle

For $x^{2} + y^{2} = 25$ , find $\frac{d^{2} y}{d x^{2}}$ in terms of $y$ .

Differentiating: $2 x + 2 y \frac{d y}{d x} = 0$ , so $\frac{d y}{d x} = - \frac{x}{y}$ . Now differentiate that with the quotient rule:

\frac{d^{2} y}{d x^{2}} = - \frac{y \cdot 1 - x \frac{d y}{d x}}{y^{2}} = - \frac{y - x (- \frac{x}{y})}{y^{2}} = - \frac{\frac{y^{2} + x^{2}}{y}}{y^{2}} = - \frac{x^{2} + y^{2}}{y^{3}} = - \frac{25}{y^{3}},

using the original equation at the last step. Substituting $\frac{d y}{d x}$ back in is the step candidates forget — leaving $\frac{d y}{d x}$ in the final answer scores nothing.

✎ Worked example — route 2 with a product term

The curve $x^{2} + x y + y^{2} = 3$ passes through $(1,1)$ . Find $\frac{d y}{d x}$ and $\frac{d^{2} y}{d x^{2}}$ there.

Differentiate once, writing $y^{'}$ for $\frac{d y}{d x}$ and using the product rule on $x y$ :

2 x + y + x y^{'} + 2 y y^{'} = 0 . (*)

At $(1,1)$ : $2 + 1 + y^{'} + 2 y^{'} = 0$ , so $y^{'} = - 1$ .

Now differentiate $(*)$ again — without rearranging it:

2 + y^{'} + (y^{'} + x y^{''}) + 2 ((y^{'})^{2} + y y^{''}) = 0 ⟹ 2 + 2 y^{'} + 2 (y^{'})^{2} + (x + 2 y) y^{''} = 0 .

At $(1,1)$ with $y^{'} = - 1$ : $2 - 2 + 2 + 3 y^{''} = 0$ , so $\frac{d^{2} y}{d x^{2}} = - \frac{2}{3}$ .

No quotient rule anywhere. When a question asks for the second derivative at a point, this route is almost always faster.

★ A* tip — the syllabus's own trick

Sometimes a rearrangement makes repeated differentiation trivial. For $y = \tan x$ ,

\frac{d y}{d x} = \sec^{2} x = 1 + \tan^{2} x = 1 + y^{2} .

Now every further derivative is pure algebra: $y^{''} = 2 y y^{'}$ , then $y^{'''} = 2 (y^{'})^{2} + 2 y y^{''}$ , and so on — no trigonometry at all. This is exactly how the Maclaurin series of $\tan x$ is built in $§ 5$ .

§4 The second derivative when $x$ and $y$ are parametric

◆ The formula — and the trap inside it

\frac{d y}{d x} = \frac{d y / d t}{d x / d t}, \frac{d^{2} y}{d x^{2}} = \frac{d}{d x} (\frac{d y}{d x}) = \frac{\frac{d}{d t} (\frac{d y}{d x})}{\frac{d x}{d t}}

Differentiate $\frac{d y}{d x}$ with respect to $t$ , then divide by $\frac{d x}{d t}$ . The chain rule demands it: $\frac{d}{d x} = \frac{d}{d t} \times \frac{d t}{d x}$ , and $\frac{d t}{d x}$ is the reciprocal of $\frac{d x}{d t}$ .

⚠ It is not

\frac{d^{2} y / d t^{2}}{d^{2} x / d t^{2}}

That looks plausible and is wrong. Test it on $x = t^{2}, y = t^{3}$ : the fake formula gives $\frac{6 t}{2} = 3 t$ , while the true answer is $\frac{3}{4 t}$ — not even the same shape. Differentiating a quotient is not the quotient of the derivatives.

✎ Worked example

For $x = t^{2}, y = t^{3}$ find $\frac{d^{2} y}{d x^{2}}$ .

$\frac{d x}{d t} = 2 t$ and $\frac{d y}{d t} = 3 t^{2}$ , so $\frac{d y}{d x} = \frac{3 t^{2}}{2 t} = \frac{3 t}{2}$ . Then

\frac{d^{2} y}{d x^{2}} = \frac{\frac{d}{d t} (\frac{3 t}{2})}{\frac{d x}{d t}} = \frac{\frac{3}{2}}{2 t} = \frac{3}{4 t} .

Check by eliminating $t$ : $y^{2} = x^{3}$ , so for $t > 0$ , $y = x^{3 / 2}$ and $\frac{d^{2} y}{d x^{2}} = \frac{3}{4} x^{- 1 / 2} = \frac{3}{4 t}$ ✓ since $x^{1 / 2} = t$ .

✎ Worked example — a messier parameter

For $x = t + \frac{1}{t}, y = t - \frac{1}{t}$ , show that $\frac{d^{2} y}{d x^{2}} = - \frac{4 t^{3}}{(t^{2} - 1)^{3}}$ .

$\frac{d x}{d t} = 1 - \frac{1}{t^{2}} = \frac{t^{2} - 1}{t^{2}}$ and $\frac{d y}{d t} = 1 + \frac{1}{t^{2}} = \frac{t^{2} + 1}{t^{2}}$ , so

\frac{d y}{d x} = \frac{t^{2} + 1}{t^{2} - 1} .

By the quotient rule, $\frac{d}{d t} (\frac{t^{2} + 1}{t^{2} - 1}) = \frac{2 t (t^{2} - 1) - (t^{2} + 1) (2 t)}{(t^{2} - 1)^{2}} = \frac{- 4 t}{(t^{2} - 1)^{2}}$ . Dividing by $\frac{d x}{d t}$ means multiplying by $\frac{t^{2}}{t^{2} - 1}$ :

\frac{d^{2} y}{d x^{2}} = \frac{- 4 t}{(t^{2} - 1)^{2}} \times \frac{t^{2}}{t^{2} - 1} = - \frac{4 t^{3}}{(t^{2} - 1)^{3}} . ■

§5 Maclaurin's series

◆ The series MF19

f (x) = f (0) + x f^{'} (0) + \frac{x^{2}}{2!} f^{''} (0) + \dots + \frac{x^{r}}{r!} f^{(r)} (0) + \dots

Each coefficient is a derivative evaluated at zero, divided by a factorial. MF19 also prints the standard expansions of $e^{x}$ , $\ln (1 + x)$ , $\sin x$ , $\cos x$ , $\tan^{- 1} x$ , $\sinh x$ , $\cosh x$ and $\tanh^{- 1} x$ , with their ranges of validity — so quote those rather than re-deriving them.

◇ The method for a series you are not given

Differentiate repeatedly. If the derivatives get ugly, look for a rearrangement that expresses $y^{'}$ in terms of $y$ — then keep differentiating that.
Substitute $x = 0$ into each derivative in turn.
Assemble $f (0) + f^{'} (0) x + \frac{f^{''} (0)}{2!} x^{2} + \dots$ , stopping at the power the question asks for.

You are never asked for the general term — only "the first few terms" or "as far as $x^{n}$ ".

✎ Worked example —

\tan x

as far as

x^{5}

Direct differentiation of $\sec^{2} x$ gets horrible fast, so use $y^{'} = 1 + y^{2}$ from $§ 3$ and differentiate that repeatedly:

y^{'} = 1 + y^{2}, y^{''} = 2 y y^{'}, y^{'''} = 2 (y^{'})^{2} + 2 y y^{''}, y^{(4)} = 6 y^{'} y^{''} + 2 y y^{'''}, y^{(5)} = 6 (y^{''})^{2} + 8 y^{'} y^{'''} + 2 y y^{(4)} .

Now evaluate at $x = 0$ , where $y = \tan 0 = 0$ , carrying each value forward:

$y (0)$	$y^{'} (0)$	$y^{''} (0)$	$y^{'''} (0)$	$y^{(4)} (0)$	$y^{(5)} (0)$
$0$	$1 + 0 = 1$	$2 (0) (1) = 0$	$2 (1)^{2} + 0 = 2$	$6 (1) (0) + 0 = 0$	$0 + 8 (1) (2) + 0 = 16$

\tan x = x + \frac{2}{3!} x^{3} + \frac{16}{5!} x^{5} + \dots = x + \frac{x^{3}}{3} + \frac{2 x^{5}}{15} + \dots

The even derivatives all vanish at $0$ , which they must: $\tan$ is an odd function, so only odd powers can appear. That parity check is free and catches a slip immediately.

✎ Worked example —

\ln (1 + \sin x)

as far as

x^{3}

Let $y = \ln (1 + \sin x)$ , so $y (0) = \ln 1 = 0$ . Then

y^{'} = \frac{\cos x}{1 + \sin x} ⟹ y^{'} (0) = \frac{1}{1} = 1 .

By the quotient rule, and using $\sin^{2} x + \cos^{2} x = 1$ to simplify the numerator:

y^{''} = \frac{- \sin x (1 + \sin x) - \cos^{2} x}{(1 + \sin x)^{2}} = \frac{- \sin x - \sin^{2} x - \cos^{2} x}{(1 + \sin x)^{2}} = \frac{- (1 + \sin x)}{(1 + \sin x)^{2}} = - \frac{1}{1 + \sin x},

so $y^{''} (0) = - 1$ . That tidy form makes the next step easy:

y^{'''} = \frac{\cos x}{(1 + \sin x)^{2}} ⟹ y^{'''} (0) = 1 .

Hence

\ln (1 + \sin x) = 0 + x - \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + \dots = x - \frac{x^{2}}{2} + \frac{x^{3}}{6} + \dots

Numerical check at $x = 0.1$ : the series gives $0.1 - 0.005 + 0.000167 = 0.095167$ , and $\ln (1 + \sin 0.1) = 0.095161$ . ✓ The tiny gap is the $x^{4}$ term.

★ A* tip — simplify each derivative before differentiating again

In the example above, spotting that $y^{''}$ collapses to $- \frac{1}{1 + \sin x}$ turned a fearsome third differentiation into a one-liner. If a derivative looks unmanageable, it is usually a signal that an identity will collapse it — look for $\sin^{2} + \cos^{2}$ , $\cosh^{2} - \sinh^{2}$ , or a common factor.

§6 Interactive: watch a Maclaurin series close in

Add terms one at a time and see the polynomial grip the curve near the origin, then peel away. The strip underneath checks every derivative formula in $§ 1$ and $§ 2$ against a numerical derivative — independent evidence that the table is right.

Include terms up to

x^{n}

Evaluate at

x

0.60

the function Maclaurin polynomial

Every derivative formula above, checked numerically

add another term ✎

§7 Examiner traps & A* checklist

⚠ The eight most common mark-losers

Writing $\frac{d}{d x} \cosh x = - \sinh x$ . There is no minus — that is the trigonometric version.
Using $\frac{d^{2} y / d t^{2}}{d^{2} x / d t^{2}}$ for a parametric second derivative. You must divide $\frac{d}{d t} (\frac{d y}{d x})$ by $\frac{d x}{d t}$ .
Leaving $\frac{d y}{d x}$ in an answer for $\frac{d^{2} y}{d x^{2}}$ — substitute it back in.
Forgetting the factorials in Maclaurin. The coefficient of $x^{r}$ is $\frac{f^{(r)} (0)}{r!}$ , not $f^{(r)} (0)$ .
Evaluating the derivatives at the general $x$ instead of at $0$ .
Mixing up $\frac{1}{\sqrt{1 - x^{2}}}$ , $\frac{1}{\sqrt{1 + x^{2}}}$ and $\frac{1}{\sqrt{x^{2} - 1}}$ — they are all on MF19, so look them up rather than guessing.
Producing even powers in the series of an odd function (or vice versa). Check the parity before you write the answer down.
Dropping the product rule on a term like $x y$ when differentiating implicitly.

★ Before you turn the page

Does my series have the right parity, and does substituting a small $x$ roughly reproduce the function?
Did I look the derivative up on MF19 rather than trusting memory?
For an implicit second derivative at a point, did I try differentiating the equation twice instead of quotient-ruling?

§8 Video explainers (curated)

▶Inverse trig & hyperbolic derivatives
Curated search — deriving them by implicit differentiation.↗ ▶Second derivatives: implicit & parametric
Curated search — the correct parametric formula, worked.↗ ▶Maclaurin's series
Curated search — successive differentiation and the standard results.↗

§9 Check yourself

Score 0 / 5

Q1 · hyperbolic derivative

$\frac{d}{d x} (\cosh x)$ equals:

Solution

$\frac{d}{d x} \frac{e^{x} + e^{- x}}{2} = \frac{e^{x} - e^{- x}}{2} = \sinh x$ . The minus belongs to $\cos$ , not $\cosh$ — which is why $y = \cosh x$ satisfies $y^{''} = y$ rather than $y^{''} = - y$ .

Q2 · inverse hyperbolic derivative

$\frac{d}{d x} (\sinh^{- 1} x)$ equals:

Solution

From $x = \sinh y$ : $1 = \cosh y y^{'}$ , so $y^{'} = \frac{1}{\cosh y} = \frac{1}{\sqrt{1 + \sinh^{2} y}} = \frac{1}{\sqrt{1 + x^{2}}}$ . The distractors are $\frac{d}{d x} \sin^{- 1} x$ , $\frac{d}{d x} \cosh^{- 1} x$ and $\frac{d}{d x} \tan^{- 1} x$ — all four are printed on MF19, so look rather than guess.

Q3 · parametric second derivative

For $x = t^{2}, y = t^{3}$ , the value of $\frac{d^{2} y}{d x^{2}}$ is:

Solution

$\frac{d y}{d x} = \frac{3 t^{2}}{2 t} = \frac{3 t}{2}$ (which is the third option — the first derivative, a common misread). Then $\frac{d^{2} y}{d x^{2}} = \frac{3 / 2}{2 t} = \frac{3}{4 t}$ . The last option, $3 t$ , is what the incorrect $\frac{d^{2} y / d t^{2}}{d^{2} x / d t^{2}} = \frac{6 t}{2}$ would give.

Q4 · Maclaurin

The Maclaurin series of $\tan x$ , as far as the term in $x^{3}$ , is:

Solution

With $y^{'} = 1 + y^{2}$ : $y (0) = 0, y^{'} (0) = 1, y^{''} (0) = 0, y^{'''} (0) = 2$ , so the $x^{3}$ coefficient is $\frac{2}{3!} = \frac{1}{3}$ . Sanity check at $x = 0.5$ : $\tan 0.5 = 0.5463$ and $0.5 + \frac{0.125}{3} = 0.5417$ — close, and correctly above $x$ . The option $x - \frac{x^{3}}{6}$ is $\sin x$ .

Q5 · full method — try it on paper first

Let $y = \ln (1 + \sin x)$ .

(i) Show that $\frac{d^{2} y}{d x^{2}} = - \frac{1}{1 + \sin x}$ .

(ii) Hence find the Maclaurin series for $\ln (1 + \sin x)$ as far as the term in $x^{3}$ .

Solution

(i) By the chain rule, $\frac{d y}{d x} = \frac{\cos x}{1 + \sin x}$ . Differentiating again with the quotient rule:

\frac{d^{2} y}{d x^{2}} = \frac{(- \sin x) (1 + \sin x) - (\cos x) (\cos x)}{(1 + \sin x)^{2}} = \frac{- \sin x - \sin^{2} x - \cos^{2} x}{(1 + \sin x)^{2}} .

Since $\sin^{2} x + \cos^{2} x = 1$ , the numerator is $- \sin x - 1 = - (1 + \sin x)$ , and one factor cancels:

\frac{d^{2} y}{d x^{2}} = \frac{- (1 + \sin x)}{(1 + \sin x)^{2}} = - \frac{1}{1 + \sin x} . ■

(ii) Part (i) makes the third derivative easy — differentiate $- (1 + \sin x)^{- 1}$ :

\frac{d^{3} y}{d x^{3}} = (1 + \sin x)^{- 2} \cos x = \frac{\cos x}{(1 + \sin x)^{2}} .

Now evaluate everything at $x = 0$ , where $\sin 0 = 0$ and $\cos 0 = 1$ :

y (0) = \ln 1 = 0, y^{'} (0) = \frac{1}{1} = 1, y^{''} (0) = - \frac{1}{1} = - 1, y^{'''} (0) = \frac{1}{1} = 1 .

Substituting into $f (0) + f^{'} (0) x + \frac{f^{''} (0)}{2!} x^{2} + \frac{f^{'''} (0)}{3!} x^{3}$ :

\ln (1 + \sin x) = x - \frac{x^{2}}{2} + \frac{x^{3}}{6} + \dots

Check at $x = 0.1$ : the series gives $0.1 - 0.005 + 0.0001667 = 0.0951667$ , and $\ln (1 + \sin 0.1) = 0.0951611$ — agreeing to five decimal places, with the discrepancy the size of the missing $x^{4}$ term. ✓ $■$

FP2 · Topic 10 · Differentiation

Differentiation Extended

What the examiner expects you to do

§1 Differentiating hyperbolic functions

§2 Inverse trigonometric and inverse hyperbolic derivatives

§3 The second derivative when y is implicit

§4 The second derivative when x and y are parametric

§5 Maclaurin's series

§6 Interactive: watch a Maclaurin series close in

Every derivative formula above, checked numerically

§7 Examiner traps & A* checklist

§8 Video explainers (curated)

§9 Check yourself

§3 The second derivative when $y$ is implicit

§4 The second derivative when $x$ and $y$ are parametric