Edexcel A-Level Mathematics A2 Fp3 3 Differentiation Questions

Practise differentiating inverse trigonometric and hyperbolic functions, simplifying exact derivatives and solving stationary-point equations.

Exam points

prove dy/dx for inverse hyperbolic forms using the chain rule and domain restrictions
determine exact roots of f'(x)=0 after simplifying square-root and rational terms
find constants in derivatives rewritten as k tan x or k sech(2x)tanh(2x)

Covered syllabus points

FP3.3.1 - Differentiation of hyperbolic cosh2x FP3.3.2 - Differentiation of inverse functions,

[Maximum number: 5]

\mathrm{g}(x)=\arctan \left(\mathrm{e}^{2 x}\right)

Show that

\mathrm{g}^{\prime \prime}(x)=k \operatorname{sech}(2 x) \tanh (2 x)

where k is a constant to be found.

$g(x)=\arctan(e^{2x})$ $g'(x)=\frac{2e^{2x}}{e^{4x}+1}=\operatorname{sech}(2x)$ $g''(x)=-2\operatorname{sech}(2x)\tanh(2x)$
So k=-2.

[Maximum number: 7]

$\quad y=\operatorname{arsinh}\left(\sqrt{x^{2}-1}\right) \quad x>1$

(a)

Prove that

\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{1}{\sqrt{x^2-1}}

where

y=\operatorname{arsinh}\left(\sqrt{x^2-1}\right),\quad x>1.

[ 3 ]

\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{1}{\sqrt{1+\left(\sqrt{x^2-1}\right)^2}}\cdot \frac{x}{\sqrt{x^2-1}}

=\frac{1}{\sqrt{x^2}}\cdot\frac{x}{\sqrt{x^2-1}}

Since x>1, $\sqrt{x^2}=x$ , so

\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{1}{\sqrt{x^2-1}}

M1: differentiates arsinh(sqrt(x^2 - 1)) using the chain rule.
A1: fully correct unsimplified derivative.
A1*: correct completion to the required result with no errors.
Other official methods using hyperbolic identities or logarithmic form are accepted.

(b)

Determine the exact values of x for which f'(x)=0.

[ 4 ]

f'(x)=\frac{1}{3\sqrt{x^2-1}}-\frac{1}{1+x^2}

f'(x)=0\Rightarrow 1+x^2=3\sqrt{x^2-1}

(1+x^2)^2=9(x^2-1)

x^4-7x^2+10=0

(x^2-2)(x^2-5)=0

x=\sqrt{2},\ \sqrt{5}

M1: forms a correct derivative equation with denominator 1 + x^2 for arctan x.
M1: cross-multiplies and squares correctly.
ddM1: solves the three-term quadratic in x^2.
A1: both exact positive solutions only.

[Maximum number: 4]

y=\operatorname{artanh}\left(\frac{\cos x+a}{\cos x-a}\right)

where a is a non-zero constant.

Show that

\frac{dy}{dx}=k\tan x

where k is a constant to be determined.

\frac{dy}{dx}=\frac{1}{1-\left(\frac{\cos x+a}{\cos x-a}\right)^2}\cdot\frac{(\cos x-a)(-\sin x)-(\cos x+a)(-\sin x)}{(\cos x-a)^2}

=\frac{(\cos x-a)^2}{(\cos x-a)^2-(\cos x+a)^2}\cdot\frac{2a\sin x}{(\cos x-a)^2}

=\frac{2a\sin x}{-4a\cos x}=-\frac12\tan x

Thus $k=-\frac12$ .

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Covered syllabus points

Question 1

Question 2

Question (a)

Question (b)

Question 3