Page 3 - Calculus

2
2
3
2
2
3
= 2( + 2 ℎ + ℎ + ℎ + 2 ℎ + ℎ )
2
3
3
=2( + 3 ℎ + 3 ℎ + ℎ )
2
=2 + 6 ℎ + 6 ℎ + 2ℎ
3
2
3
2

′
( ) = lim ( +ℎ)− ( )
ℎ→0 ℎ

3
3
2
2
2 +6 ℎ+6 ℎ +2ℎ − 2 3
= lim
ℎ→0 ℎ

2
2
6 ℎ+6 ℎ +2ℎ 3
= lim
ℎ→0 ℎ
2
ℎ(6 ²+6 ℎ+2ℎ )
= lim
ℎ→0 ℎ
2
2
= 6 + 6 (0) + 2(0)
2
( )= 6
′

Summary
• The derivative, ′( ) is a gradient fuction of ( ). Thus it represents the gradient of the ( ) at
any point.

• we know that gradient, = y 1 −y 2
x 1 −x 2

now consider P[ ; ( )] and Q[ + ℎ; ( + ℎ)]
which lie on ( )

Average gradient between P and Q
m = 1 − 2 = ( +ℎ)− ( ) = ( +ℎ)− ( )
1 − 2 +ℎ− h
but h is very small, in other words h is approaching 0 thus ℎ → 0. Therefore by definition (from
the first principles), the gradient function is given by

′
( )= lim ( +ℎ)− ( )
ℎ→0 ℎ