CAIE A-Level Further Math A2 2.3 Differentiation Questions

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: 4]

Figure for Question 1 — CAIE A-Level Further Math A2

The diagram shows part of the curve $y=x \operatorname{sech}^{2} x$ and its maximum point M.

Show that, at M,

2 x \tanh x-1=0

and verify that this equation has a root between 0.7 and 0.8 .

$\begin{aligned} \frac{\mathrm{d} y}{\mathrm{~d} x}=-2 x \operatorname{sech}^{2} x \tanh x+\operatorname{sech}^{2} x {\left[=1-\tanh ^{2} x-2 x \tanh x+2 x \tanh ^{3} x\right]} \end{aligned}$
Differentiating using product rule
and chain rule for M1.

$\operatorname{sech}^{2} x \neq 0 \Rightarrow 2 x \tanh x-1=0$
AG.

$\begin{aligned} 2(0.7) \tanh (0.7)-1=-0.154<0 2(0.8) \tanh (0.8)-1=0.062>0 \end{aligned}$
Shows sign change. Must write
down values correct to at least 1dp
for B 1 .

4

[Maximum number: 5]

The curve C has parametric equations

x=\frac{2}{3} t^{\frac{3}{2}}-2 t^{\frac{1}{2}}, \quad y=2 t+5, \quad \text { for } 0<t \leqslant 3 .

Find the set of values of t for which $\frac{\mathrm{d}^{2} y}{\mathrm{~d} x^{2}}>0$ .

$\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{\dot{y}}{\dot{x}}=\frac{2}{t^{\frac{1}{2}}-t^{-\frac{1}{2}}}$
Finds first derivative.

$\frac{\mathrm{d}}{\mathrm{d} t}\left(\frac{2}{t^{\frac{1}{2}}-t^{-\frac{1}{2}}}\right)=\frac{-2\left(\frac{1}{2} t^{-\frac{1}{2}}+\frac{1}{2} t^{-\frac{3}{2}}\right)}{\left(t^{\frac{1}{2}}-t^{-\frac{1}{2}}\right)^{2}}$
Differentiates $\frac{\mathrm{d} y}{\mathrm{~d} x}$ with respect to t.

$\frac{\mathrm{d}^{2} y}{\mathrm{~d} x^{2}}=\frac{\mathrm{d}}{\mathrm{d} t}\left(\frac{2}{t^{\frac{1}{2}}-t^{-\frac{1}{2}}}\right) \times \frac{\mathrm{d} t}{\mathrm{~d} x}=\frac{-2\left(\frac{1}{2} t^{-\frac{1}{2}}+\frac{1}{2} t^{-\frac{3}{2}}\right)}{\left(t^{\frac{1}{2}}-t^{-\frac{1}{2}}\right)^{3}}=-\frac{t+1}{(t-1)^{3}}$
Applies chain rule. OE. Does not have to be simplified
for A1.

0<t<1
Accept -1<t<1. CWO.

5

[Maximum number: 6]

Find the first three terms in the Maclaurin's series for $\tanh ^{-1}\left(\frac{1}{2} \mathrm{e}^{x}\right)$ in the form $\frac{1}{2} \ln a+b x+c x^{2}$ , giving the exact values of the constants a, b and c.

$\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{\frac{1}{2} \mathrm{e}^{x}}{1-\frac{1}{4} \mathrm{e}^{2 x}}$ $\frac{\mathrm{d}^{2} y}{\mathrm{~d} x^{2}}=\frac{\left(1-\frac{1}{4} \mathrm{e}^{2 x}\right)\left(\frac{1}{2} \mathrm{e}^{x}\right)-\frac{1}{2} \mathrm{e}^{x}\left(-\frac{1}{2} \mathrm{e}^{2 x}\right)}{\left(1-\frac{1}{4} \mathrm{e}^{2 x}\right)^{2}}$ $f^{\prime}(0)=\frac{2}{3} \quad f^{\prime \prime}(0)=\frac{10}{9}$
Evaluates derivatives at x=0.

$f(0)=\tanh ^{-1}\left(\frac{1}{2}\right)=\frac{1}{2} \ln \left(\frac{3}{2} \times 2\right)$
Uses logarithmic form of $\tanh ^{-1}$ .

$\frac{1}{2} \ln 3+\frac{2}{3} x+\frac{5}{9} x^{2}$
Applies $f(x)=f(0)+f^{\prime}(0) x+\frac{1}{2!} f^{\prime \prime}(0) x^{2}$

6

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Covered syllabus points

Question 1

Question 2

Question 3