Past exam of the mathematics course of the University of Cambridge / 2024 / ia / Paper 1 / 10E / b

Table of contents
- i b
  - Solution i
- ii b
  - Solution ii

i

Solution

Write

f (a + h) g (a + h) - f (a) g (a) = [f (a + h) - f (a)] g (a + h) + f (a) [g (a + h) - g (a)] .

Divide by

h

and let

h \to 0

. Differentiability implies continuity, so

g (a + h) \to g (a)

. Therefore the product rule gives

(f g)^{'} (a) = f^{'} (a) g (a) + f (a) g^{'} (a) .

ii

Solution

No. Take

a = 0

and

f (x) = x, g (x) = 1 + ∣ x ∣.

Then

f

is differentiable and

(f g) (x) = x + x ∣ x ∣.

Since

x ∣ x ∣/ x = ∣ x ∣ \to 0

, the product is differentiable at zero with

(f g)^{'} (0) = 1 \neq = 0.

However,

g

has one-sided derivatives

- 1

and

1

and is not differentiable at zero. This is a nondifferentiable factor with a differentiable nondegenerate product.

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