CAIE A-Level Further Mathematics 2.3 Differentiation

Practise differentiating inverse hyperbolic, implicit and parametric relations, then using derivatives at zero to construct exact Maclaurin expansions.

Exam points

apply hyperbolic identities with product and chain rules to simplify an exact derivative
for parametric curves, divide first derivatives and apply the parameter chain rule again for curvature
evaluate successive derivatives at zero and insert the factorial factors in the Maclaurin formula

Covered syllabus points

2.3.1—Hyperbolic functions 2.3.2—An expression for x y d 2.3.3—Maclaurin series

[Maximum number: 3]

Show that $\frac{\mathrm{d} y}{\mathrm{~d} x}=-\operatorname{cosech} x$ .

$\begin{aligned} \frac{\mathrm{d} y}{\mathrm{~d} x}=-\frac{\operatorname{sech}^{2}\left(\frac{1}{2} x\right)}{2 \tanh \left(\frac{1}{2} x\right)}=-\frac{1}{2 \sinh \left(\frac{1}{2} x\right) \cosh \left(\frac{1}{2} x\right)} \text { Or } \frac{\mathrm{d} y}{\mathrm{~d} x}=-\frac{\operatorname{cosech}^{2}\left(\frac{1}{2} x\right)}{2 \operatorname{coth}\left(\frac{1}{2} x\right)}=-\frac{1}{2 \sinh \left(\frac{1}{2} x\right) \cosh \left(\frac{1}{2} x\right)} \end{aligned}$

M1 A1

Uses chain rule.

$=-\frac{1}{\sinh (x)}=-\operatorname{cosech} x$

A1

AG

3

[Maximum number: 7]

It is given that

x=1+\frac{1}{t} \quad \text { and } \quad y=t \mathrm{e}^{t} .

(a)

Show that $\frac{\mathrm{d} y}{\mathrm{~d} x}=-\mathrm{e}^{t}\left(t^{3}+t^{2}\right)$ .

[ 3 ]

$\frac{\mathrm{d} y}{\mathrm{~d} t}=\mathrm{e}^{t}(t+1), \frac{\mathrm{d} x}{\mathrm{~d} t}=-t^{-2}$

B1

$\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{\mathrm{d} y}{\mathrm{dt}} \times \frac{\mathrm{dt}}{\mathrm{d} x}=-t^{2} \mathrm{e}^{t}(t+1)=-\mathrm{e}^{t}\left(t^{3}+t^{2}\right)$

M1 A1

Uses chain rule, AG.

3

(b)

Find $\frac{\mathrm{d}^{2} y}{\mathrm{~d} x^{2}}$ in terms of t.

[ 4 ]

$\frac{\mathrm{d}}{\mathrm{d} t}\left(\frac{\mathrm{~d} y}{\mathrm{~d} x}\right)=-\mathrm{e}^{t}\left(3 t^{2}+2 t\right)-\mathrm{e}^{t}\left(t^{3}+t^{2}\right)=-\mathrm{e}^{t}\left(t^{3}+4 t^{2}+2 t\right)$

M1 A1

Applies product rule.

$\frac{\mathrm{d}^{2} y}{\mathrm{~d} x^{2}}=t^{3} \mathrm{e}^{t}\left(t^{2}+4 t+2\right)$

M1 A1

Uses chain rule, multiplies by their
$\frac{\mathrm{d} t}{\mathrm{~d} x}$ for M 1 .

Alternative method for question 2(b)

$\frac{\mathrm{d}^{2} y}{\mathrm{~d} x^{2}}=-\mathrm{e}^{t}\left(3 t^{2} \frac{\mathrm{~d} t}{\mathrm{~d} x}+2 t \frac{\mathrm{~d} t}{\mathrm{~d} x}\right)-\mathrm{e}^{t}\left(t^{3}+t^{2}\right) \frac{\mathrm{d} t}{\mathrm{~d} x}$

(M1 A1)

Differentiates $-\mathrm{e}^{t}\left(t^{3}+t^{2}\right)$
implicitly.

$\frac{\mathrm{d}^{2} y}{\mathrm{~d} x^{2}}=\mathrm{e}^{t}\left(-4 t^{2}-2 t-t^{3}\right) \frac{\mathrm{d} t}{\mathrm{~d} x}=t^{3} \mathrm{e}^{t}\left(t^{2}+4 t+2\right)$

(M1 A1)

Substitutes $\frac{\mathrm{d} t}{\mathrm{~d} x}=-t^{2}$ .

4

[Maximum number: 9]

(a)

Given that $y=\tanh ^{-1}\left(\frac{1-x}{2+x}\right)$ , show that $(2 x+1) \frac{\mathrm{d} y}{\mathrm{~d} x}+1=0$ .

[ 4 ]

EITHER Solution 1
$y=\frac{1}{2} \ln \left(\frac{1+\frac{1-x}{2+x}}{1-\frac{1-x}{2+x}}\right)$

1

(M1

Use result of (a)

$=\frac{1}{2} \ln \left(\frac{3}{2 x+1}\right)$

1

A1

Differentiate: $y^{\prime}=\frac{1}{2} \times \frac{2 x+1}{3} \times \frac{-6}{(2 x+1)^{2}}$

1

M1

Any equivalent work, e.g. splitting the log into
2 terms

$=\frac{-1}{2 x+1}$ giving $(2 x+1) \frac{\mathrm{d} y}{\mathrm{~d} x}+1=0$

1

A1)

AG

$\begin{aligned} \text { OR Solution 2 } \text { Differentiate first: } \operatorname{sech}^{2} y \frac{\mathrm{~d} y}{\mathrm{~d} x}=\frac{-3}{(2+x)^{2}} \end{aligned}$

2

(M1A1

Use $\operatorname{sech}^{2} y=1-\tanh ^{2} y$

1

M1

to give $(2 x+1) \frac{\mathrm{d} y}{\mathrm{~d} x}+1=0$

1

A1)

AG

Available marks

4

(b)

Hence find the first three terms in the Maclaurin's series for $\tanh ^{-1}\left(\frac{1-x}{2+x}\right)$ in the form

a \ln 3+b x+c x^{2},

where a, b and c are constants to be determined.

[ 5 ]

Differentiate again: $y^{\prime \prime}=\frac{2}{(2 x+1)^{2}}$

1

B1

Or differentiate result in (b): $(2 x+1) y^{\prime \prime}+2 y^{\prime}=0$ $y(0)=\tanh ^{-1}\left(\frac{1}{2}\right)$

1

M1

Values of derivatives at 0

$y^{\prime}(0)=-1 ; y^{\prime \prime}(0)=2$

1

M1

Use Maclaurin's series

$y=\frac{1}{2} \ln 3-x+2 \times \frac{x^{2}}{2}$

1

M1

In terms of ln

$=\frac{1}{2} \ln 3-x+x^{2}$

1

A1

CAO

5

All question bank results loaded

Covered syllabus points

2.3 Differentiation question 1

2.3 Differentiation question 2

Question (a)

Question (b)

2.3 Differentiation question 3

Question (a)

Question (b)