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Mathematics Term 2 STPM Chapter 2 Differentiation
= lim 2xx + (x) 2 4
x → 0 x
= lim (2x + x)
x → 0
= 2x
1 _ 1
d 1 x + x x
(b) 1 2 = lim 4
dx x x → 0 x
= lim x – (x + x) 4
x → 0 (x + x)x (x)
2
= lim –1 4
x → 0 x(x + x)
= – 1
x 2
Example 2
dy
2
If y = 3x + x + 1, find from the first principles.
dx
2
2
dy 3(x + x) + (x + x) + 1 – (3x + x + 1)
Solution: = lim 4
dx x → 0 x
2
2
2
= lim 3x + 6xx + 3(x) + x + x + 1 – 3x – x – 1 4
x → 0 x
2
= lim 6x x + 3(x) + x 4
x → 0 x
= lim [6x + 3 x + 1]
x → 0
= 6x + 1
Exercise 2.1
Find the derivatives of the following functions from the first principles.
3
4
1. y = x 2. y = x 3. y = 5x 2
2
2
4. y = 1 5. y = x + 5x 6. y = x – x + 3
x 2
2
3
7. y = 4x 8. y = 1 9. y = 2x + 3
x 3
10. y = 2x – 3x + 1
2
18
02 STPM Math(T) T2.indd 18 02/11/2018 12:43 PM

