Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2009 / Paper 1 / 11 / a

Table of contents
- i a
  - Solution i
- ii a
  - Solution ii

i

Solution

Differentiate the numerator and denominator using the product rule and quotient rule:

y^{'} = \frac{( 4 x ^{3} + 3 x ^{2} ) ( 3 x - 2 ) - 12 x ^{3} ( x + 1 )}{( 3 x - 2 ) ^{5}} .

The numerator simplifies to

- 11 x^{3} - 6 x^{2}

y^{'} = - \frac{x ^{2} ( 11 x + 6 )}{( 3 x - 2 ) ^{5}}, x \neq = \frac{2}{3} .

ii

Solution

For the real inverse hyperbolic tangent, take

a \neq = 0

and

∣ x / a ∣ < 1

. Its logarithmic expression gives

y = \frac{1}{2} lo g \frac{1 + x / a}{1 - x / a}, y^{'} = \frac{a}{a ^{2} - x ^{2}} .

The same derivative holds on a differentiable complex branch away from its singularities.

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