The document discusses the concept of differentiability using limits and four examples: 1) The function f(x) = x - 5 is not differentiable at x = 5 since the left and right hand derivatives are not equal. 2) The function f(x) = x^x, 0 ≤ x < 2 and f(x) = (x-1)x, 2 ≤ x < 3 is not differentiable at x = 2 since the left and right hand derivatives are not equal. 3) The function f(x) = x^2, x ≤ 1 and f(x) = 1/x, x > 1 is not differentiable at x = 1 since the left and

1. DIFFERENTIATION USING
FIRST PRINCIPLES

2. A function f is said to be derivable at x = c, if
lim
ℎ→0+
𝑓 𝑐 + ℎ − 𝑓(𝑐)
ℎ
= lim
ℎ→0−
𝑓 𝑐 + ℎ − 𝑓(𝑐)
ℎ
Right hand derivative Left hand derivative
Or
lim
ℎ→0
𝑓 𝑐 + ℎ − 𝑓(𝑐)
ℎ
exists and is denoted by 𝑓′
(𝑐)

3. Question 1
Discuss the differentiability of f(x) = 𝑥 − 5 at x = 5
Right hand derivative = lim
ℎ→0+
𝑓 5+ℎ −𝑓(5)
ℎ
=
= lim
ℎ→0+
ℎ
ℎ
= lim
ℎ→0
ℎ
ℎ
= 1
= lim
ℎ→0+
5 + ℎ − 5 − 5 − 5
ℎ

4. Left hand derivative = lim
ℎ→0−
𝑓 5+ℎ −𝑓(5)
ℎ
= lim
ℎ→0−
5+ℎ−5 − 5−5
ℎ
Since right hand derivative ≠ 𝑙𝑒𝑓𝑡 ℎ𝑎𝑛𝑑 𝑑𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑣𝑒, 𝑓 𝑖𝑠 𝑛𝑜𝑡 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡𝑖𝑎𝑏𝑙𝑒
at x = 5
= lim
ℎ→0−
ℎ
ℎ
= lim
ℎ→0
−ℎ
ℎ
= −1

5. Question 2
Examine the function f defined by 𝑓 𝑥 = 𝑥 𝑥 , 0 ≤ 𝑥 < 2
= (x-1)x. 2≤ 𝑥 < 3
for differentiability at x = 2
Left hand derivative = lim
ℎ→0−
𝑓 2+ℎ −𝑓(2)
ℎ
= lim
ℎ→0
2+ℎ 2+ℎ −2
ℎ
= lim
ℎ→0
2 + ℎ − 2
ℎ
= lim
ℎ→0
ℎ
ℎ
= 1

6. Right hand derivative
lim
ℎ→0+
𝑓 2 + ℎ − 𝑓(2)
ℎ
= lim
ℎ→0
2 + ℎ − 1 2 + ℎ − 2
ℎ
=lim
ℎ→0
1+ℎ 2+ℎ −2
ℎ
= lim
ℎ→0
2+3ℎ+ℎ2−2
ℎ
= lim
ℎ→0
ℎ(ℎ + 3)
ℎ
= lim
ℎ→0
ℎ + 3 = 3

7. Left hand derivative ≠ right hand derivative
The function is not differentiable at x = 2
Question 3
𝑓 𝑥 = 𝑥2
, 𝑥 ≤ 1
=
1
𝑥
, x >1
Discuss the differentiability of f at x = 1

8. lim
ℎ→0+
𝑓 1 + ℎ − 𝑓(1)
ℎ
= lim
ℎ→0
1
1 + ℎ
− 1
ℎ
Right hand derivative
= lim
ℎ→0
1 − (1 + ℎ)
ℎ(1 + ℎ)
= lim
ℎ→0
−ℎ
ℎ(1 + ℎ)
= −1

9. Left hand derivative
lim
ℎ→0−
𝑓 1 + ℎ − 𝑓(1)
ℎ
= lim
ℎ→0
(1 + ℎ)2−1
ℎ
= lim
ℎ→0
1 + 2ℎ + ℎ2 − 1
ℎ
= lim
ℎ→0
ℎ + 2 =2
Right hand derivative ≠ 𝑙𝑒𝑓𝑡 ℎ𝑎𝑛𝑑 𝑑𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑣𝑒
f is not differentiable at x = 1

10. Question 4
Find the right hand and left hand derivatives at x = 1. Hence find the condition for the existence of the
derivative at x = 1
𝑓 𝑥 = 𝑎𝑥2
+ 1, 𝑥 ≥ 1
= x +a, x <1
lim
ℎ→0+
𝑓 1 + ℎ − 𝑓(1)
ℎ
= lim
ℎ→0
𝑎(1 + ℎ)2
+ 1 − (𝑎 + 1)
ℎ
= lim
ℎ→0
𝑎 1 + 2ℎ + ℎ2
+ 1 − 𝑎 − 1
ℎ
Right hand derivative

11. = lim
ℎ→0
𝑎 + 2𝑎ℎ + 𝑎ℎ2
+ 1 − 𝑎 − 1
ℎ
= lim
ℎ→0
2𝑎 + 𝑎ℎ
=2a
Left hand derivative
lim
ℎ→0−
𝑓 1 + ℎ − 𝑓(1)
ℎ
= lim
ℎ→0
1 + ℎ + 𝑎 − (𝑎 + 1)
ℎ
= lim
ℎ→0
1 + ℎ + 𝑎 − 𝑎 − 1
ℎ
= 1

12. 2𝑎 = 1
Since f is derivable at x = 1
𝑎 =
1
2
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