Chapter 11 and 12: Differentiation Rules and Proofs
Mr RJ Maartens
These are the differentiation rules and proofs that you must know and be able to apply.
1. Constant Rule (Page 500)
If 𝑓(𝑥) = 𝑐, where 𝑐 is any real number, then 𝑓 ′ (𝑥) = 0.
Proof
𝑓(𝑥 + ℎ) − 𝑓(𝑥)
ℎ→0
ℎ
𝑐−𝑐
= lim
ℎ→0 ℎ
0
= lim
ℎ→0 ℎ
= lim 0
𝑓 ′ (𝑥) = lim
ℎ→0
=0
2. Power Rule (Page 501)
If 𝑓(𝑥) = 𝑥 𝑛 , where 𝑛 is any real number, then 𝑓 ′ (𝑥) = 𝑛𝑥 𝑛−1 .
Proof
Leave out this proof.
3. Constant Product Rule (Page 503)
If 𝑓(𝑥) = 𝑐. 𝑔(𝑥), where 𝑐 is any real number, then 𝑓 ′ (𝑥) = 𝑐. 𝑔′ (𝑥).
Proof
𝑓(𝑥 + ℎ) − 𝑓(𝑥)
ℎ→0
ℎ
𝑐. 𝑔(𝑥 + ℎ) − 𝑐. 𝑔(𝑥)
= lim
ℎ→0
ℎ
𝑐[𝑔(𝑥 + ℎ) − 𝑔(𝑥)]
= lim
ℎ→0
ℎ
𝑔(𝑥 + ℎ) − 𝑔(𝑥)
= lim 𝑐.
ℎ→0
ℎ
𝑔(𝑥 + ℎ) − 𝑔(𝑥)
= 𝑐. lim
ℎ→0
ℎ
= 𝑐. 𝑔′ (𝑥)
𝑓 ′ (𝑥) = lim
4. Sum Rule (Page 504)
If 𝑓(𝑥) = 𝑚(𝑥) + 𝑛(𝑥), then 𝑓 ′ (𝑥) = 𝑚′ (𝑥) + 𝑛′ (𝑥).
Proof
𝑓(𝑥 + ℎ) − 𝑓(𝑥)
ℎ→0
ℎ
[𝑚(𝑥 + ℎ) + 𝑛(𝑥 + ℎ)] − [𝑚(𝑥) + 𝑛(𝑥)]
= lim
ℎ→0
ℎ
𝑚(𝑥 + ℎ) + 𝑛(𝑥 + ℎ) − 𝑚(𝑥) − 𝑛(𝑥)
= lim
ℎ→0
ℎ
[𝑚(𝑥 + ℎ) − 𝑚(𝑥)] + [𝑛(𝑥 + ℎ) − 𝑛(𝑥)]
= lim
ℎ→0
ℎ
[𝑚(𝑥 + ℎ) − 𝑚(𝑥)] [𝑛(𝑥 + ℎ) − 𝑛(𝑥)]
= lim
+
ℎ→0
ℎ
ℎ
[𝑚(𝑥 + ℎ) − 𝑚(𝑥)]
[𝑛(𝑥 + ℎ) − 𝑛(𝑥)]
= lim
+ lim
ℎ→0
ℎ→0
ℎ
ℎ
′ (𝑥)
′ (𝑥)
=𝑚
+𝑛
𝑓 ′ (𝑥) = lim
5. Difference Rule (Page 504)
If 𝑓(𝑥) = 𝑚(𝑥) − 𝑛(𝑥), then 𝑓 ′ (𝑥) = 𝑚′ (𝑥) − 𝑛′ (𝑥).
Proof
𝑓(𝑥 + ℎ) − 𝑓(𝑥)
ℎ→0
ℎ
[𝑚(𝑥 + ℎ) − 𝑛(𝑥 + ℎ)] − [𝑚(𝑥) − 𝑛(𝑥)]
= lim
ℎ→0
ℎ
𝑚(𝑥 + ℎ) − 𝑛(𝑥 + ℎ) − 𝑚(𝑥) + 𝑛(𝑥)
= lim
ℎ→0
ℎ
[𝑚(𝑥 + ℎ) − 𝑚(𝑥)] − [𝑛(𝑥 + ℎ) − 𝑛(𝑥)]
= lim
ℎ→0
ℎ
[𝑚(𝑥 + ℎ) − 𝑚(𝑥)] [𝑛(𝑥 + ℎ) − 𝑛(𝑥)]
= lim
−
ℎ→0
ℎ
ℎ
[𝑚(𝑥 + ℎ) − 𝑚(𝑥)]
[𝑛(𝑥 + ℎ) − 𝑛(𝑥)]
= lim
− lim
ℎ→0
ℎ→0
ℎ
ℎ
′ (𝑥)
′ (𝑥)
=𝑚
−𝑛
𝑓 ′ (𝑥) = lim
6. Product Rule (Page 518)
If 𝑓(𝑥) = 𝑚(𝑥). 𝑛(𝑥), then 𝑓 ′ (𝑥) = 𝑛(𝑥). 𝑚′ (𝑥) + 𝑚(𝑥). 𝑛′ (𝑥).
Alternatively, if 𝑢 = 𝑚(𝑥), 𝑣 = 𝑛(𝑥) and 𝑓(𝑥) = 𝑢𝑣, then 𝑓 ′ (𝑥) = 𝑣𝑢′ + 𝑢𝑣′.
Proof
𝑓(𝑥 + ℎ) − 𝑓(𝑥)
ℎ→0
ℎ
[𝑚(𝑥 + ℎ). 𝑛(𝑥 + ℎ)] − [𝑚(𝑥). 𝑛(𝑥)]
= lim
ℎ→0
ℎ
𝑚(𝑥 + ℎ). 𝑛(𝑥 + ℎ) − 𝑚(𝑥). 𝑛(𝑥) + 𝒎(𝒙). 𝒏(𝒙 + 𝒉) − 𝒎(𝒙). 𝒏(𝒙 + 𝒉)
= lim
ℎ→0
ℎ
𝑚(𝑥 + ℎ). 𝑛(𝑥 + ℎ) − 𝒎(𝒙). 𝒏(𝒙 + 𝒉) + 𝒎(𝒙). 𝒏(𝒙 + 𝒉) − 𝑚(𝑥). 𝑛(𝑥)
= lim
ℎ→0
ℎ
𝑛(𝑥 + ℎ)[𝑚(𝑥 + ℎ) − 𝑚(𝑥)] + 𝑚(𝑥)[𝑛(𝑥 + ℎ) − 𝑛(𝑥)]
= lim
ℎ→0
ℎ
𝑚(𝑥 + ℎ) − 𝑚(𝑥)
𝑛(𝑥 + ℎ) − 𝑛(𝑥)
= lim 𝑛(𝑥 + ℎ) [
] + lim 𝑚(𝑥) [
]
ℎ→0
ℎ→0
ℎ
ℎ
𝑚(𝑥 + ℎ) − 𝑚(𝑥)
𝑛(𝑥 + ℎ) − 𝑛(𝑥)
= [lim 𝑛(𝑥 + ℎ)] [lim
] + [lim 𝑚(𝑥)] [lim
]
ℎ→0
ℎ→0
ℎ→0
ℎ→0
ℎ
ℎ
= 𝑛(𝑥 + 0). 𝑚′ (𝑥) + 𝑚(𝑥). 𝑛′ (𝑥)
= 𝑛(𝑥). 𝑚′ (𝑥) + 𝑚(𝑥). 𝑛′ (𝑥)
𝑓 ′ (𝑥) = lim
7. Quotient Rule (Page 521)
If 𝑓(𝑥) =
𝑚(𝑥)
𝑛(𝑥)
, where 𝑛(𝑥) ≠ 0, then 𝑓 ′ (𝑥) =
𝑛(𝑥).𝑚′ (𝑥)− 𝑚(𝑥).𝑛′ (𝑥)
.
[𝑛(𝑥)]2
𝑢
Alternatively, if 𝑢 = 𝑚(𝑥), 𝑣 = 𝑛(𝑥) and 𝑓(𝑥) = 𝑣 , where 𝑣 ≠ 0, then 𝑓 ′ (𝑥) =
𝑣𝑢′ −𝑢𝑣 ′
𝑣2
.
Proof
𝑓(𝑥 + ℎ) − 𝑓(𝑥)
ℎ→0
ℎ
1
= lim [𝑓(𝑥 + ℎ) − 𝑓(𝑥)]
ℎ→0 ℎ
1 𝑚(𝑥 + ℎ) 𝑚(𝑥)
= lim [
−
]
ℎ→0 ℎ 𝑛(𝑥 + ℎ)
𝑛(𝑥)
1 𝑚(𝑥 + ℎ)𝑛(𝑥) − 𝑚(𝑥)𝑛(𝑥 + ℎ)
= lim [
]
ℎ→0 ℎ
𝑛(𝑥 + ℎ)𝑛(𝑥)
𝑚(𝑥 + ℎ)𝑛(𝑥) − 𝑚(𝑥)𝑛(𝑥 + ℎ)
= lim
ℎ→0
ℎ. 𝑛(𝑥 + ℎ)𝑛(𝑥)
𝑚(𝑥 + ℎ)𝑛(𝑥) − 𝑚(𝑥)𝑛(𝑥 + ℎ) + 𝒎(𝒙)𝒏(𝒙) − 𝒎(𝒙)𝒏(𝒙)
= lim
ℎ→0
ℎ. 𝑛(𝑥 + ℎ)𝑛(𝑥)
𝑚(𝑥 + ℎ)𝑛(𝑥) − 𝒎(𝒙)𝒏(𝒙) − 𝑚(𝑥)𝑛(𝑥 + ℎ) + 𝒎(𝒙)𝒏(𝒙)
= lim
ℎ→0
ℎ. 𝑛(𝑥 + ℎ)𝑛(𝑥)
𝑛(𝑥)[𝑚(𝑥 + ℎ) − 𝑚(𝑥)] − 𝑚(𝑥)[𝑛(𝑥 + ℎ) + 𝑛(𝑥)]
= lim
ℎ→0
𝒉. 𝑛(𝑥 + ℎ)𝑛(𝑥)
𝑚(𝑥 + ℎ) − 𝑚(𝑥)
𝑛(𝑥 + ℎ) − 𝑛(𝑥)
𝑛(𝑥) [
] − 𝑚(𝑥) [
]
𝒉
𝒉
= 𝐥𝐢𝐦
𝒉→𝟎
𝑛(𝑥 + ℎ)𝑛(𝑥)
𝑚(𝑥 + ℎ) − 𝑚(𝑥)
𝑛(𝑥 + ℎ) − 𝑛(𝑥)
𝑛(𝑥) [𝐥𝐢𝐦
] − 𝑚(𝑥) [𝐥𝐢𝐦
]
ℎ
ℎ
𝒉→𝟎
𝒉→𝟎
=
𝐥𝐢𝐦 𝑛(𝑥 + ℎ) 𝑛(𝑥)
𝑓 ′ (𝑥) = lim
𝒉→𝟎
𝑛(𝑥)𝑚′ (𝑥) − 𝑚(𝑥)𝑛′ (𝑥)
=
𝑛(𝑥 + 0)𝑛(𝑥)
𝑛(𝑥)𝑚′ (𝑥) − 𝑚(𝑥)𝑛′ (𝑥)
=
[𝑛(𝑥)]2
You can also use the alternative proof on Page 521, however, then you must include the proof
of the Product rule here. See comment at the bottom of the page.
8. Chain Rule (Page 527)
If 𝑓(𝑥) = 𝑔(ℎ(𝑥)), then 𝑓 ′ (𝑥) = 𝑔′ (ℎ(𝑥)). ℎ′ (𝑥).
𝑑𝑓
𝑑𝑓
𝑑𝑢
Alternatively, if 𝑢 = ℎ(𝑥) and 𝑓(𝑥) = 𝑔(𝑢), then 𝑑𝑥 = 𝑑𝑢 × 𝑑𝑥 .
Proof
Leave out this proof.
9. 𝒍𝒏-Rule (Page 541)
1
If 𝑓(𝑥) = ln(𝑥), then 𝑓 ′ (𝑥) = 𝑥 for 𝑥 > 0.
Proof
𝑓(𝑥 + ℎ) − 𝑓(𝑥)
ℎ→0
ℎ
1
= lim [𝑓(𝑥 + ℎ) − 𝑓(𝑥)]
ℎ→0 ℎ
1
= lim [ln(𝑥 + ℎ) − ln(𝑥)]
ℎ→0 ℎ
1
𝑥+ℎ
= lim . ln (
)
ℎ→0 ℎ
𝑥
1
ℎ
= lim . ln (1 + )
ℎ→0 ℎ
𝑥
1 𝒙
ℎ
= lim [ . ] ln (1 + )
ℎ→0 𝒙 ℎ
𝑥
1
𝒙
ℎ
= . lim ln (1 + )
𝑥 ℎ→0 𝒉
𝑥
𝒙
1
ℎ ⁄𝒉
= . lim ln (1 + )
𝑥 ℎ→0
𝑥
𝑥
1
ℎ ⁄ℎ
= . lim ln (1 + )
𝑥 𝒉⁄𝒙→0
𝑥
1
𝟏
= . lim ln(1 + 𝒌) ⁄𝒌
𝑥 𝒌→0
1
1
= . ln [𝐥𝐢𝐦(1 + 𝑘) ⁄𝑘 ]
𝒌→𝟎
𝑥
1
= . ln[𝑒]
𝑥
1
=
𝑥
𝑓 ′ (𝑥) = lim
*Note that 𝑒 = lim (1 + 𝑘)
𝑘→0
1⁄
𝑘,
see page 467.
10. 𝒍𝒐𝒈-Rule (Page 543)
𝑢′
1
If 𝑓(𝑥) = log 𝑏 𝑢, where 𝑢 = 𝑔(𝑥), then 𝑓 ′ (𝑥) = ln(𝑏) .
𝑢
for 𝑢 > 0.
Proof
𝑑
[log 𝑏 𝑢]
𝑑𝑥
𝑑 ln(𝑢)
=
[
]
𝑑𝑥 ln(𝑏)
1
𝑑
=
.
ln(𝑢)
ln(𝑏) 𝑑𝑥
1
𝑢′
=
.
ln(𝑏) 𝑢
𝑓 ′ (𝑥) =
*Note that the chain rule is used here for 𝑢.
11. Inverse Function Rule (Page 545)
𝑑
1
If 𝑓(𝑥) is an invertible, differentiable function, then 𝑑𝑥 𝑓 −1 (𝑥) = 𝑓′ (𝑓−1 (𝑥)).
𝑑
1
Alternatively, if 𝑘 = 𝑓 −1 (𝑥) for some invertible, differentiable 𝑓(𝑥), then 𝑑𝑥 𝑘 = 𝑓′ (𝑘).
Proof
Since,
𝑑
𝑑
𝑥=
𝑥
𝑑𝑥
𝑑𝑥
𝑑
𝑑
𝑓(𝑓 −1 (𝑥)) =
𝑥
𝑑𝑥
𝑑𝑥
We can use the chain rule on the right-hand side, and the power rule on the left:
𝑓 ′ (𝑓 −1 (𝑥)).
∴
𝑑𝑦
𝑑 −1
𝑓 (𝑥) = 1
𝑑𝑥
𝑑 −1
1
𝑓 (𝑥) = ′ −1
𝑑𝑥
𝑓 (𝑓 (𝑥))
*Note that this rule implies that 𝑑𝑥 =
1
𝑑𝑥
( )
𝑑𝑦
𝑑𝑥
and 𝑑𝑦 =
1
.
𝑑𝑦
( )
𝑑𝑥
12. 𝒆-Rule (Page 546)
If 𝑓(𝑥) = 𝑒 𝑥 , then 𝑓 ′ (𝑥) = 𝑒 𝑥 .
Proof
If 𝑦 = 𝑓(𝑥) = 𝑒 𝑥 , we have:
𝑦 = 𝑒𝑥
∴ 𝑥 = ln(𝑦)
∴
∴
𝑑𝑥 1
1
= = 𝑥
𝑑𝑦 𝑦 𝑒
𝑑𝑦
1
1
=
=
= 𝑒𝑥
𝑑𝑥 (𝑑𝑥 ) ( 1 )
𝑒𝑥
𝑑𝑦
from the inverse function rule.
13. Exponential Rule (Page 547)
If 𝑓(𝑥) = 𝑏 𝑢 , where 𝑏 is any real number and 𝑢 = 𝑔(𝑥), then 𝑓 ′ (𝑥) = 𝑏 𝑢 × ln(𝑏) × 𝑢′ .
Proof
Using the chain rule:
𝑑 𝑢
𝑏
𝑑𝑥
𝑑 ln(𝑏𝑢)
=
𝑒
𝑑𝑥
𝑑
𝑢
= 𝑒 ln(𝑏 ) . ln(𝑏 𝑢 )
𝑑𝑥
𝑑
𝑢
= 𝑒 ln(𝑏 ) . [𝑢. ln(𝑏)]
𝑑𝑥
𝑑
𝑢
= 𝑒 ln(𝑏 ) . ln(𝑏) . [𝑢]
𝑑𝑥
= 𝑏 𝑢 . ln(𝑏) . 𝑢′
𝑓 ′ (𝑥) =