Cambridge International AS and A Level Mathematics Pure Mathematics 1

Differentiation

P1^

5

KEY POINTS

1   y = kxn ⇒ d

d

y

x

=knxn–1

y = c ⇒ d

d

y

x=^0

2  y = f(x) + g(x) ⇒ d

d

y

x

= f′(x) + g′(x).

3  Tangent and normal at (x 1 , y 1 )

Gradient of tangent, m 1 = value of d

d

y

x

when x = x 1.

Gradient of normal, m 2 = –^1

m 1

.

Equation of tangent is

y − y 1 = m 1 (x − x 1 ).

Equation of normal is

y − y 1 = m 2 (x − x 1 ).

4  At a stationary point, d

d

y

x

= 0.

The nature of a stationary point can be determined by looking at the sign of

the gradient just either side of it.

5  The nature of a stationary point can also be determined by considering the

sign of d

d

2

2

y

x

.

● If

d

d

2

2

y

x

< 0, the point is a maximum.

● If

d

d

2

2

y

x

> 0, the point is a minimum.

6  If

d

d

2

2

y

x

= 0, check the values of d

d

y

x

on either side of the point to determine

its nature.

7  Chain rule: d

d

d

d

d

d

y

x

y

u

u

x

=×.

Where k, n and c are

constants.

}

0

+ –

0

Maximum Minimum Stationary point of infection

0

+

+

+

• 0
  
  
  • 
    
  
  
  
  

+