Calculus 3-The Chain Rule, guaranteed 100% Pass

1


The Chain Rule


The Chain Rule:
𝑑𝑦 𝑑𝑦 𝑑𝑢
In 1 variable if 𝑦 = 𝑓(𝑢), 𝑢 = 𝑔(𝑥), then = 𝑑𝑢 .
𝑑𝑥 𝑑𝑥

Ex. 𝑦 = 𝑢10 ; 𝑢 = sin 𝑥 (i.e. 𝑦 = (sin 𝑥 )10 ) then
𝑑𝑦
= 10𝑢9 (cos 𝑥) = 10(sin 𝑥)9 cos 𝑥.
𝑑𝑥



Ex. 𝑧 = 𝑢10 ; 𝑢(𝑥, 𝑦) = sin 𝑥𝑦 then 𝑧 = sin10 𝑥𝑦.


𝜕𝑧 9
= (10(sin 𝑥𝑦)) (cos 𝑥𝑦)𝑦
𝜕𝑥

𝜕𝑧
= 10(sin 𝑥𝑦)9 (cos 𝑥𝑦)𝑥.
𝜕𝑦



For functions of more than 1 variable, the chain rule can take several forms
depending on the situation.



Case 1: 𝑧 = 𝑓 (𝑥, 𝑦) ; 𝑥 = 𝑥 (𝑡) ; 𝑦 = 𝑦(𝑡)


𝑑𝑧 𝜕𝑓 𝑑𝑥 𝜕𝑓 𝑑𝑦 𝜕𝑧 𝑑𝑥 𝜕𝑧 𝑑𝑦
= + = +
𝑑𝑡 𝜕𝑥 𝑑𝑡 𝜕𝑦 𝑑𝑡 𝜕𝑥 𝑑𝑡 𝜕𝑦 𝑑𝑡

, 2

𝑑𝑧 𝜋
Ex. Let 𝑧 = 𝑥 4 + 𝑦 4 ; and 𝑥 = 𝑠𝑖𝑛𝑡, 𝑦 = 𝑐𝑜𝑠𝑡. Find at 𝑡 = .
𝑑𝑡 6

𝑑𝑧 𝜕𝑧 𝑑𝑥 𝜕𝑧 𝑑𝑦
= 𝜕𝑥 + 𝜕𝑦
𝑑𝑡 𝑑𝑡 𝑑𝑡

𝜕𝑧 𝜕𝑧 𝑑𝑥 𝑑𝑦
= 4𝑥 3 , = 4𝑦 3 , = 𝑐𝑜𝑠𝑡, = −𝑠𝑖𝑛𝑡.
𝜕𝑥 𝜕𝑦 𝑑𝑡 𝑑𝑡

𝑑𝑧
= 4𝑥 3 (cos 𝑡) + 4𝑦 3 (− sin 𝑡) (∗)
𝑑𝑡
= 4 sin3 𝑡 (cos 𝑡) − 4 cos3 𝑡 (sin 𝑡) .
𝜋
At 𝑡 = we get:
6

𝑑𝑧 𝜋 𝜋 𝜋 𝜋
| 𝜋 = 4 sin3 (cos ) − 4 (cos3 )(sin )
𝑑𝑡 𝑡= 6 6 6 6 6
3
1 3 √3 √3 1
= 4( ) ( ) − 4( ) ( )
2 2 2 2

4 √3 12√3 8 √3 √3
= − =− =−
16 16 16 2

𝜋
Note: When we got to (∗), we could have said when 𝑡 = ,
6
𝜋 1 𝜋 √3
𝑥 = sin = , and 𝑦 = cos = and then plugged in
6 2 6 2

3
𝑑𝑧 1 3 √3 √3 1
= 4 ( 2) ( 2 ) − 4 ( 2 ) (2)
𝑑𝑡

4√3 12√3 8√3 √3
= 16
− 16
=− 16
=− 2
.
𝑑𝑧
However, if we had been asked to find at any point 𝑡
𝑑𝑡
we would have needed to substitute 𝑥 = 𝑠𝑖𝑛𝑡, 𝑦 = 𝑐𝑜𝑠𝑡
into (∗).