Cambridge International AS and A Level Mathematics Pure Mathematics 1

P1^

5

Exercise

(^) 5B
EXAMPLE 5.4 Differentiate f(x) = ()xx()
x
(^2) +− 15
SOLUTION
You cannot differentiate f(x) as it stands, so you need to start by rewriting it.
Expanding the brackets:
Now you can differentiate f(x) to give f′(x) = 2 x − 5 + 5 x−^2
= 2 x + (^52)
x

− 5

EXERCISE 5B Differentiate the following functions using the rules

y = kxn ⇒ d

d

y

x

= knxn−^1

and y = f(x) + g(x) ⇒ d

d

y

x

= f′(x) + g′(x).

  1  y = x^5    2 y = 4 x^2   3 y = 2 x^3

(^4) y = x^11   5 y = 4 x^10 6 y = 3 x^5
  7  y = 7 8 y = 7 x   9 y = 2 x^3 + 3 x^5
10  y = x^7 − x^4 11 y = x^2 + 1 12 y = x^3 + 3 x^2 + 3 x + 1
13  y = x^3 − 9 14 y = 12 x^2 + x + 1 15 y = 3 x^2 + 6 x + 6
16  A = 4 πr^2 17 A =^43 πr^3 18 d =^14 t^2
19  C = 2 πr 20 V = l^3 21 f(xx)=
(^32)
22 y
x
=^1 23 yx= 24 yx=^15
(^52)
25 f(x)
x
=^12 26 f(x)
x

=^53

(^27) y
x
=^2
28 f(xx)
x
=− 4 8
29 f(xx)=+x−
(^3232)
30 f(xx)=−x−
(^5323)
31 y = x(4x − 1) 32 f(x) = (2x − 1)(x + 3) 33 y xx
= x
(^2) + (^6)
(^34) y xx
x
=^45 −
64
2 35 yx= x^36 f(x)
x
x
=^2
37 g(x) xx
x
=^32 −
2
38 y x
x
=+()xx−
4
(^4) () 2
39 h(xx)=()
3
(^40) y xx x
x
=()+−()
(^224)
2
f(x) xx x
x
x
x
x
x
x
xx
xx x
= −+ −
=− +−
=− +− −
32
32
2 1
55
55
51 5