Page 36 - Focus SPM 2022 - Additional Mathematics

Additional Mathematics SPM Chapter 2 Differentiation

3 Therefore, inductively,
dy
n
Given y = 2x , find dy for n = 1, 2 and 3. dx = 2 = 1(2x 1 – 1 )
dx dy
Hence, verify the first derivative formula dy = nax n – 1 dx = 4x = 2(2x 2 – 1 )
inductively. dx dy 2 3 – 1
dx = 6x = 3(2x )
formula.Rights Reserved.
Solution
(a) When n = 1, dy = n(ax n – 1 ) = nax n – 1
y = 2x dx
y + dy = 2(x + dx)
2x + dy = 2x + 2dx B Determining the first derivative of
an algebraic function
dy = 2dx
dy = 2 1. For any function f(x) or y written in the
form of ax where a and n are constants, f'(x) or
n
dx dy n – 1
= ax
.
lim dy
dx
\ dx → 0 dx = 2 2. When determining the first derivative of an
and dy = 2 algebraic functions, make sure that each term
dx is written in the form of ax before applying the
n
When n = 2,
y = 2x 2 3. dy or f'(x) is known as a gradient function and it
dx
y + dy = 2(x + dx) 2 is used to determine the gradient of the tangent
2
2x + dy = 2[x + 2xdx + (dx) ] to a curve of y or f(x) at a point on the curve.
2
2
2x + dy = 2x + 4xdx + 2(dx)
2
2
2
dy = 4xdx + 2x(dx) 4. In summary:
2
• function y = a (a = constant)
dy = dx[4x + 2xdx] dy
dy = 4x + 2xdx dx = 0
dx • function y = ax (a, n = constant)
n
lim dy
dy

Penerbitan Pelangi Sdn Bhd. All dx = nax n – 1
= 4x
\
dx → 0
dx
n
m
and dy = 4x • function y = ax + bx + …
dx (a, n, b, m = constant)
dy
When n = 3, dx = nax n – 1 + mbx m – 1 + …
y = 2x 3
y + dy = 2(x + dx) 3 4 Form 5
2x + dy = 2[x + 3x dx + 3x(dx) + (dx) ] Find the first derivative of each of the following
3
3
2
3
2
2x + dy = 2x + 6x dx + 6x(dx) + 2(dx) function.
3
3
2
3
2
dy = 6x dx + 6x(dx) + 2(dx) (a) y = 6x
2
3
2
3
dy = 6x + 6xdx + 2x(dx) (b) y = 5
2
2
4x
2
dx (c) y = 3√x
lim dy
\ dx → 0 dx = 6x 2 Solution
3
and dy = 6x (a) y = 6x
2
dx dy = 3[6x 3 – 1 ]
dx
2
dy = 18x
dx
245