BEngineering Mathematics 1 (MAT 2018) July 2020

Unit 2 Partial Differentiation

2.2 Partial Derivatives

In engineering, functions are generally multivariate (or multivariable); that is, they have more
than one input (or independent) variable. For example, to calculate volume (or cubic capacity)
of a cylinder the distance between the circular faces and the area of each face is required.

Varying either (or both) the height and radius of the face will increase/ decrease the volume, or
it may remain unchanged.

The Partial Derivative of a function of several variables is its derivative with respect to one of
those variables, with the others held constant. Partial derivatives are used in vector calculus and
differential geometry.

Note: For this module, MAT 2018, the focus is on bivariate functions, that is, functions with
two input variables.

2.2.1 Definition of Partial Derivative (Bivariate Function)

Given the bivariate function, 𝑧 = 𝑓(𝑥, 𝑦), the first-order partial derivative with respect to x is
𝝏𝒛 𝒇(𝒙+𝒉,𝒚)−𝒇(𝒙,𝒚)
defined by 𝒛𝒙 = = 𝐥𝐢𝐦 .
𝝏𝒙 𝒉 →𝟎 𝒉

Example Given 𝑧 = 4𝑥 − 3𝑦, deduce and expression for 𝑧𝑥 .

𝜕𝑧
Solution 𝑧𝑥 𝑂𝑅 are commonly used symbols for the partial derivative with respect to x.
𝜕𝑥
1
 Essentially, we are differentiating with respect to x, while treating the variable y as
though it was a constant (just a number).

𝜕 𝜕 𝜕
(4𝑥 − 3𝑦) = (4𝑥) − (3𝑦) = 4 − 0
𝜕𝑥 𝜕𝑥 𝜕𝑥
𝜕
Note: (3𝑦) = 0, since y is treated like a constant.
𝜕𝑥

Example Given 𝑧 = 9𝑥 3 + 3𝑦 2 + 7𝑥𝑦, deduce an expression for 𝑧𝑥 .

𝜕 𝜕 𝜕
Solution 𝑧𝑥 = (9𝑥 3 ) + (3𝑦 2 ) + (7𝑥𝑦) = 27𝑥 2 + 0 + 7𝑦 = 27𝑥 2 + 7𝑦
𝜕𝑥 𝜕𝑥 𝜕𝑥

𝜕 𝜕
Note: (7𝑥𝑦) = (7𝑦𝑥) = 7𝑦; 7𝑦 is the coefficient of x.
𝜕𝑥 𝜕𝑥

Similarly, Given the bivariate function, 𝑧 = 𝑓(𝑥, 𝑦), the first-order partial derivative with
𝝏𝒛 𝒇(𝒙,𝒚+𝒉)−𝒇(𝒙,𝒚)
respect to y is defined by 𝒛𝒚 = = 𝐥𝐢𝐦 .
𝝏𝒚 𝒉 →𝟎 𝒉

Example Given 𝑧 = 𝑒 2𝑥−3𝑦 , deduce expressions for 𝑧𝑥 and 𝑧𝑦 .

Solution 𝑧𝑥 = 2𝑒 2𝑥−3𝑦 and 𝑧𝑦 = −3𝑒 2𝑥−3𝑦

Example Given 𝑧 = 3sin (𝑥 2 + 5𝑥𝑦), deduce expressions for 𝑧𝑥 and 𝑧𝑦 .

Solution 𝑧𝑥 = 3cos (𝑥 2 + 5𝑥𝑦) × (2𝑥 + 5𝑦) and 𝑧𝑦 = 3cos (𝑥 2 + 5𝑥𝑦) × (5𝑥)

Example Given 𝑧 = 𝑥 2 ln (4𝑦 − 5𝑥), deduce expressions for 𝑧𝑥 and 𝑧𝑦 .

−5 4
Solution 𝑧𝑥 = 𝑥 2 ( ) + 2𝑥𝑙𝑛(4𝑦 − 5𝑥); using product rule and 𝑧𝑦 = 𝑥 2 ( )
4𝑦−5𝑥 4𝑦−5𝑥

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2.2.2 Finding the Partial Derivatives (Worked Examples)

Example For each of the following, find the partial derivatives, 𝑧𝑥 and 𝑧𝑦 :

a) 𝑧 = 7𝑥 − 6𝑦 2 + 4𝑥 3 𝑦 5 e) 𝑧 = 4𝑒 7𝑥𝑦 − ln (𝑥𝑦)
b) 𝑧 = (3𝑥 − 4𝑦 + 5)7 f) 𝑧 = 𝑥 𝑦 + 𝑦 𝑥
2 +𝑦 3
c) 𝑧 = 4𝑠𝑖𝑛(3𝑥𝑦) − 7𝑐𝑜𝑠(𝑥 2 + 3𝑦 3 ) g) 𝑧 = 𝑒 𝑥
d) 𝑧 = 𝑥 𝑠𝑖𝑛(𝑥𝑦) − 𝑦 𝑐𝑜𝑠(𝑥𝑦) h) 𝑧 = 𝑥 𝑙𝑛( 𝑥𝑦)

Solution

Function, 𝒛 = 𝒇(𝒙, 𝒚) 𝒛𝒙 𝒛𝒚
a) 𝑧 = 7𝑥 − 6𝑦 2 + 4𝑥 3 𝑦 5 𝑧𝑥 = 7 + 12𝑥 2 𝑦 5 𝑧𝑦 = −12𝑦 + 20𝑥 3 𝑦 4
b) 𝑧 = (3𝑥 − 4𝑦 + 5)7 𝑧𝑥 = 7(3𝑥 − 4𝑦 + 5)6 × 3 = 𝑧𝑦 = 7(3𝑥 − 4𝑦 + 5)6 ×
21(3𝑥 − 4𝑦 + 5)6 (−4) = −28(3𝑥 − 4𝑦 + 5)6
Chain rule was applied
c) 𝑧 = 4𝑠𝑖𝑛(3𝑥𝑦) − 𝑧𝑥 = 4 × 3𝑦𝑐𝑜𝑠(3𝑥𝑦) + 𝑧𝑦 = 4 × 3𝑥𝑐𝑜𝑠(3𝑥𝑦) +
7𝑐𝑜𝑠(𝑥 2 + 3𝑦 3 ) 7 sin(𝑥 2 + 3𝑦 3 ) × 2𝑥 = 7 sin(𝑥 2 + 3𝑦 3 ) × 9𝑦 2 =
12𝑦𝑐𝑜𝑠(3𝑥𝑦) + 14𝑥𝑠𝑖𝑛(𝑥 2 + 12𝑥𝑐𝑜𝑠(3𝑥𝑦) +
3𝑦 3 ) 63𝑦 2 𝑠𝑖𝑛(𝑥 2 + 3𝑦 3 )
Chain rule was applied
d) 𝑧 = 𝑥 𝑠𝑖𝑛(𝑥𝑦) − 𝑧𝑥 = 𝑥𝑦𝑐𝑜𝑠(𝑥𝑦) + 𝑧𝑦 = 𝑥 2 cos(𝑥𝑦) −
𝑦 𝑐𝑜𝑠(𝑥𝑦) (1) sin(𝑥𝑦) + 𝑦 2 sin (𝑥𝑦) [−𝑥𝑦𝑠𝑖𝑛(𝑥𝑦) +
Product rule applied (1) cos(𝑥𝑦)] = 𝑥 2 cos(𝑥𝑦) +
𝑥𝑦𝑠𝑖𝑛(𝑥𝑦) − cos (𝑥𝑦)
Product rule applied
𝑦 𝑥
e) 𝑧 = 4𝑒 7𝑥𝑦 − ln (𝑥𝑦) 𝑧𝑥 = 4𝑒 7𝑥𝑦 (7𝑦) − = 𝑧𝑦 = 4𝑒 7𝑥𝑦 (7𝑥) − =
𝑥𝑦 𝑥𝑦
1 1
28𝑦𝑒 7𝑥𝑦 − 28𝑥𝑒 7𝑥𝑦 −
𝑥 𝑦

f) 𝑧 = 𝑥 𝑦 + 𝑦 𝑥 𝑧𝑥 = 𝑦𝑥 𝑦−1 + 𝑦 𝑥 𝑙𝑛𝑦 𝑧𝑦 = 𝑥 𝑦 𝑙𝑛𝑥 + 𝑥𝑦 𝑥−1

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 2 +𝑦 3 2 +𝑦 3 2 +𝑦 3
g) 𝑧 = 𝑒 𝑥 𝑧𝑥 = 2𝑥𝑒 𝑥 𝑧𝑦 = 3𝑦 2 𝑒 𝑥
h) 𝑧 = 𝑥 𝑙𝑛( 𝑥𝑦) 𝑦 𝑥 𝑥
𝑧𝑥 = 𝑥 ( ) + (1) ln(𝑥𝑦) = 𝑧𝑦 = 𝑥 ( )=
𝑥𝑦 𝑥𝑦 𝑦
1 + ln (𝑥𝑦) ; Product rule

𝜕𝑧 𝜕𝑧
Example Given 𝑧 = 𝑙𝑛(𝑒 𝑥 + 𝑒 𝑦 ), show that + = 1.
𝜕𝑥 𝜕𝑦

𝜕𝑧 𝑒𝑥 𝜕𝑧 𝑒𝑦
Solution = and = ; by applying the chain rule.
𝜕𝑥 𝑒 𝑥 +𝑒 𝑦 𝜕𝑦 𝑒 𝑥 +𝑒 𝑦

𝜕𝑧 𝜕𝑧 𝑒𝑥 𝑒𝑦 𝑒 𝑥 +𝑒 𝑦
𝐿𝐻𝑆: + = + = = 1 (𝑅𝐻𝑆); shown.
𝜕𝑥 𝜕𝑦 𝑒 𝑥 +𝑒 𝑦 𝑒 𝑥 +𝑒 𝑦 𝑒 𝑥 +𝑒 𝑦

𝑥2𝑦2 𝜕𝑧 𝜕𝑧
Example If 𝑍 = ; show that 𝑥 +𝑦 = 2𝑧.
𝑥 2 +𝑦 2 𝜕𝑥 𝜕𝑦

𝜕𝑧 𝜕𝑧
Solution Firstly, let us find and using the quotient rule:
𝜕𝑥 𝜕𝑦

𝜕𝑧 (𝑥 2 +𝑦 2 )2𝑥𝑦 2 −𝑥 2 𝑦 2 (2𝑥) 2𝑥 3 𝑦 2 +2𝑥𝑦 4 −2𝑥 3 𝑦 2 2𝑥𝑦 4

= (𝑥 2 +𝑦 2 )2
= (𝑥 2 +𝑦 2 )2
= (𝑥 2
𝜕𝑥 +𝑦 2 )2

𝜕𝑧 (𝑥 2 +𝑦 2 )2𝑥 2 𝑦−𝑥 2 𝑦 2 (2𝑦) 2𝑥 4 𝑦+2𝑥 2 𝑦 3 −2𝑥 2 𝑦 3 2𝑥 4 𝑦

= (𝑥 2 +𝑦 2 )2
= (𝑥 2 +𝑦 2 )2
= (𝑥 2
𝜕𝑦 +𝑦 2 )2

Now, starting with the LHS:

𝜕𝑧 𝜕𝑧 2𝑥𝑦 4 2𝑥 4 𝑦 2𝑥 2 𝑦 4 +2𝑥 4 𝑦 2 2𝑥 2 𝑦 2 (𝑦 2 +𝑥 2 ) 2𝑥 2 𝑦 2
𝑥 +𝑦 = 𝑥 ((𝑥 2 ) + 𝑦 ((𝑥 2 +𝑦2)2) = = = = 2𝑧
𝜕𝑥 𝜕𝑦 +𝑦 2 )2 (𝑥 2 +𝑦 2 )2 (𝑥 2 +𝑦 2 )2 𝑥 2 +𝑦 2

Example Find the first-order partial derivatives, 𝑧𝑥 and 𝑧𝑦 , of 𝑧 = sin−1 (5𝑥 2 𝑦 2 ).

1 10𝑥𝑦 2
Solution 𝑧𝑥 = 10𝑥𝑦 2 ( )= ; by chain rule
√1−(5𝑥 2 𝑦 2 ) 2 √1−25𝑥 4 𝑦 4

1 10𝑥 2 𝑦
𝑧𝑦 = 10𝑥 2 𝑦 ( )= ; by chain rule
√1−(5𝑥 2 𝑦 2 ) 2 √1−25𝑥 4 𝑦 4

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2.2.3 Second Order Partial Derivatives

Recall In the univariate case where 𝑦 = 𝑓(𝑥) and the first-order derivative with respect to
𝑑𝑦 𝑑2𝑦 𝑑𝑦
x is , the second-order derivative, , is obtained by differentiating with
𝑑𝑥 𝑑𝑥 2 𝑑𝑥

respect to x.

In the multivariate case, there are several second-order derivatives and one must specify “with
respect to____”

First, given a bivariate function, 𝑧 = 𝑓(𝑥, 𝑦) , the first-order partial derivatives are:

𝜕𝑓 𝜕𝑧 𝜕
𝑧𝑥 ≡ 𝑓𝑥 (𝑥, 𝑦) ≡ 𝑓𝑥 ≡ ≡ ≡ (𝑓(𝑥, 𝑦)) when the function is differentiated with respect to
𝜕𝑥 𝜕𝑥 𝜕𝑥

x. The notations will vary by users.

𝜕𝑓 𝜕𝑧 𝜕
Similarly, 𝑧𝑦 ≡ 𝑓𝑦 (𝑥, 𝑦) ≡ 𝑓𝑦 ≡ ≡ ≡ (𝑓(𝑥, 𝑦)) when the function is differentiated with
𝜕𝑦 𝜕𝑦 𝜕𝑦

respect to y.

Illustrative Example Given the bivariate function, 𝑧 = 𝑓(𝑥, 𝑦) = 9𝑥 3 + 4𝑥 2 𝑦 4 − 7𝑦 5 ,

Find the first-order partial derivatives, 𝑧𝑥 and 𝑧𝑦 .

Solution 𝑧𝑥 = 27𝑥 2 + 8𝑥𝑦 4 and 𝑧𝑦 = 16𝑥 2 𝑦 3 − 35𝑦 4

The second-order partial derivatives are as follows:

𝜕 𝜕 𝜕𝑧 𝜕2 𝑧
1) 𝑧𝑥𝑥 =
𝜕𝑥
(𝑧𝑥 ) = ( ) = 𝜕𝑥 2 = 54𝑥 + 8𝑦 4 ; here 𝑧𝑥 is differentiated with respect to x.
𝜕𝑥 𝜕𝑥
𝜕 𝜕 𝜕𝑧 𝜕2 𝑧
2) 𝑧𝑥𝑦 =
𝜕𝑦
(𝑧𝑥 ) = ( ) = 𝜕𝑦𝜕𝑥 = 32𝑥𝑦 3 ; here the partial derivative, 𝑧𝑥 , is
𝜕𝑦 𝜕𝑥

𝜕2 𝑧
differentiated with respect to y to obtain 𝑧𝑥𝑦 or . Please note the order of the
𝜕𝑦𝜕𝑥

subscripts in the notation and the denominator of the partial derivative.

𝜕 𝜕 𝜕𝑧 𝜕2 𝑧
3) 𝑧𝑦𝑦 =
𝜕𝑦
(𝑧𝑦 ) = 𝜕𝑦 (𝜕𝑦) = 𝜕𝑦2 = 48𝑥 2 𝑦 2 − 140𝑦 3
𝜕 𝜕 𝜕𝑧 𝜕2 𝑧
4) 𝑧𝑦𝑥 =
𝜕𝑥
(𝑧𝑦 ) = 𝜕𝑥 (𝜕𝑦) = 𝜕𝑥𝜕𝑦 = 32𝑥𝑦 3

5
Note 𝑧𝑥𝑦 = 𝑧𝑦𝑥 ; is not coincidental. This applies to many “nice” functions.

Clairaut’s Theorem (not on syllabus nor required)

If 𝑓(𝑥, 𝑦) is defined on a disk that contains the point (𝑎, 𝑏) and 𝑓𝑥𝑦 and 𝑓𝑦𝑥 are continuous on
the disk then 𝑓𝑥𝑦 (𝑎, 𝑏) = 𝑓𝑦𝑥 (𝑎, 𝑏)

Example Find all the second order partial derivatives of the following:

a) 𝑧 = 𝑒 3𝑥𝑦 b) 𝑧 = 3𝑠𝑖𝑛(4𝑥 − 3𝑦) c) 𝑧 = 𝑙𝑛(𝑥 2 𝑦 2 )

Solution

a) 𝑧 = 𝑒 3𝑥𝑦
First-order partials: 𝑧𝑥 = 3𝑦𝑒 3𝑥𝑦 𝑧𝑦 = 3𝑥𝑒 3𝑥𝑦
Second-order partials: 𝑧𝑥𝑥 = 9𝑦 2 𝑒 3𝑥𝑦 𝑧𝑦𝑦 = 9𝑥 2 𝑒 3𝑥𝑦

𝑧𝑥𝑦 = (3𝑦)(3𝑥𝑒 3𝑥𝑦 ) + 3𝑒 3𝑥𝑦 = 9𝑥𝑦𝑒 3𝑥𝑦 + 3𝑒 3𝑥𝑦 𝑂𝑅 3𝑒 3𝑥𝑦 (3𝑥𝑦 + 1); product rule
𝑧𝑦𝑥 = (3𝑥)(3𝑦𝑒 3𝑥𝑦 ) + 3𝑒 3𝑥𝑦 = 9𝑥𝑦𝑒 3𝑥𝑦 + 3𝑒 3𝑥𝑦 𝑂𝑅 3𝑒 3𝑥𝑦 (3𝑥𝑦 + 1); product rule

b) 𝑧 = 3𝑠𝑖𝑛(4𝑥 − 3𝑦)
First-order partials: 𝑧𝑥 = 12cos (4𝑥 − 3𝑦) 𝑧𝑦 = −9cos (4𝑥 − 3𝑦)

Second-order partials: 𝑧𝑥𝑥 = −48sin (4𝑥 − 3𝑦) 𝑧𝑦𝑦 = −27sin (4𝑥 − 3𝑦)

𝑧𝑥𝑦 = 36sin (4𝑥 − 3𝑦) 𝑧𝑦𝑥 = 36sin (4𝑥 − 3𝑦)

c) 𝑧 = 𝑙𝑛(𝑥 2 𝑦 2 )
2𝑥𝑦 2 2 2𝑥 2 𝑦 2
First-order partials: 𝑧𝑥 = = = 2𝑥 −1 𝑧𝑦 = = = 2𝑦 −1
𝑥2𝑦2 𝑥 𝑥2𝑦2 𝑦
2 2
Second-order partials: 𝑧𝑥𝑥 = −2𝑥 −2 = − 𝑧𝑦𝑦 = −2𝑦 −2 = −
𝑥2 𝑦2

𝑧𝑥𝑦 = 0 𝑧𝑦𝑥 = 0

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 𝑦 𝜕2 𝑧 𝜕2 𝑧
Example If 𝑧 = 𝑡𝑎𝑛−1 ( ), show that 2 + 2 = 0 .
𝑥 𝜕𝑥 𝜕𝑦

Solution Let us begin with finding the first and second-order partials with respect to x:

𝑦
𝑧 = 𝑡𝑎𝑛−1 ( ) = tan−1 (𝑦𝑥 −1 )
𝑥

1 −𝑦 1 −𝑦
𝑧𝑥 = −𝑦𝑥 −2 ( )= (1+𝑦2𝑥 −2) = 𝑥 2 +𝑦2 = −𝑦(𝑥 2 + 𝑦 2 )−1
1+(𝑦𝑥 −1 )2 𝑥2

2𝑥𝑦
𝑧𝑥𝑥 = 𝑦(𝑥 2 + 𝑦 2 )−2 × 2𝑥 = (𝑥 2
+𝑦 2 )2

1 1 1 1 𝑥 𝑥
𝑧𝑦 = 𝑥 −1 ( ) = 𝑥 (1+𝑦2𝑥 −2) = 𝑥+𝑦2 𝑥 −1 × 𝑥 = 𝑥 2+𝑦2 = 𝑥(𝑥 2 + 𝑦 2 )−1
1+(𝑦𝑥 −1 )2

−2𝑥𝑦
𝑧𝑦𝑦 = −𝑥(𝑥 2 + 𝑦 2 )−2 × 2𝑦 = (𝑥 2
+𝑦 2 )2

Now, let us start with the LHS:

𝜕2 𝑧 𝜕2 𝑧 2𝑥𝑦 −2𝑥𝑦
2
+ = (𝑥 2 + (𝑥 2 = 0; shown.
𝜕𝑥 𝜕𝑦 2 +𝑦 2 )2 +𝑦 2 )2

2.2.4 Chain Rule

Recall In the univariate case, the chain rule is applicable to composite functions (function
of a function), such as 𝑦 = (9 − 3𝑥 2 )4 ; in which the inside function is identified as
𝑢 = 9 − 3𝑥 2 . The derivative may then be found by applying the chain rule, as
follows:
𝑑𝑢
𝑢 = 9 − 3𝑥 2 = −6𝑥
𝑑𝑥

𝑑𝑦
𝑦 = 𝑢4 = 4𝑢3
𝑑𝑢

𝑑𝑦 𝑑𝑦 𝑑𝑢
= × = 4𝑢3 × (−6𝑥) = −24𝑥(9 − 3𝑥 2 )3
𝑑𝑥 𝑑𝑢 𝑑𝑥

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In the bivariate case, z is a function of x and y → 𝑧 = 𝑓(𝑥, 𝑦)

Then each of x and/or y are functions of other variables. Indirectly z is a function of the other
variables.

Let us consider two simple examples.

Illustrative Example 1 Let 𝑧 = 𝑓(𝑥, 𝑦) = 4𝑥 3 + 3𝑦 2 with 𝑥 = 4 + 3𝑡 and 𝑦 = 2𝑡 − 1 .

𝑧 = 𝑓(𝑥, 𝑦) = 4𝑥 3 + 3𝑦 2 may be expressed solely in terms of t, and then the

𝑑𝑧
derivative, , is found. Please note that the partial symbols were not used since z
𝑑𝑡

is univariate when expressed in terms of t only, hence it is not a partial derivative.

𝑧 = 𝑓(𝑥, 𝑦) = 4𝑥 3 + 3𝑦 2 = 4(4 + 3𝑡)3 + 3(2𝑡 − 1)2

𝑑𝑧
Then, = 36(4 + 3𝑡)2 + 12(2𝑡 − 1)
𝑑𝑡

The derivative may also be found using the Chain Rule.

𝒅𝒛 𝝏𝒛 𝒅𝒙 𝝏𝒛 𝒅𝒚
In which case, if 𝑧 = 𝑓(𝑥, 𝑦) with 𝑥 = 𝑔(𝑡) and 𝑦 = ℎ(𝑡); then = × + ×
𝒅𝒕 𝝏𝒙 𝒅𝒕 𝝏𝒚 𝒅𝒕

Let us re-do the example using the Chain rule:

𝑑𝑧
= (12𝑥 2 )(3) + (6𝑦)(2) = 36𝑥 2 + 12𝑦 = 36(4 + 3𝑡)2 + 12(2𝑡 − 1)
𝑑𝑡

Illustrative Example 2 Let 𝑧 = 𝑓(𝑥, 𝑦) = 4𝑥 3 + 3𝑦 2 with 𝑥 = 4𝑠 2 𝑡 and 𝑦 = 3𝑡 − 𝑠 2 .

In this example 𝑧 = 𝑓(𝑥, 𝑦) = 4𝑥 3 + 3𝑦 2 may be expressed in terms of s and t by

simple substitution:

𝑧 = 𝑓(𝑠, 𝑡) = 4(4𝑠 2 𝑡)3 + 3(3𝑡 − 𝑠 2 )2 = 256𝑠 6 𝑡 3 + 3(3𝑡 − 𝑠 2 )2

𝜕𝑧 𝜕𝑧
The partial derivatives, and , may then be easily found, as follows:
𝜕𝑠 𝜕𝑡

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 𝜕𝑧
= 6 × 256𝑠 5 𝑡 3 + 6 × (−2𝑠)(3𝑡 − 𝑠 2 ) = 1536𝑠 5 𝑡 3 − 12𝑠(3𝑡 − 𝑠 2 )
𝜕𝑠

𝜕𝑧
= 3(256)𝑠 6 𝑡 2 + 6(3)(3𝑡 − 𝑠 2 ) = 768𝑠 6 𝑡 2 + 18(3𝑡 − 𝑠 2 )
𝜕𝑡

This example may be represented using a tree diagram (see below):

𝜕𝑧
To find → follow the branches that end with t, and we multiply along the
𝜕𝑠

branches, then add them up. This gives the following result:

𝝏𝒛 𝝏𝒛 𝝏𝒙 𝝏𝒛 𝝏𝒚
= × + × = (12𝑥 2 )(8𝑠𝑡) + (6𝑦)(−2𝑠)
𝝏𝒔 𝝏𝒙 𝝏𝒔 𝝏𝒚 𝝏𝒔
= 96𝑠𝑡(4𝑠 2 𝑡 )2 − 12𝑠(3𝑡 − 𝑠 2 ) = 1536𝑠 5 𝑡 3 − 12𝑠(3𝑡 − 𝑠 2 )

Which is the same result that we got previously using direct substitution.

Similarly,

𝝏𝒛 𝝏𝒛 𝝏𝒙 𝝏𝒛 𝝏𝒚
= × + × = (12𝑥 2 )(4𝑠 2 ) + (6𝑦)(3) = 48𝑠 2 (4𝑠 2 𝑡 )2 − 18𝑦
𝝏𝒕 𝝏𝒙 𝝏𝒕 𝝏𝒚 𝝏𝒕
= 768𝑠 6 𝑡 2 − 18(3𝑡 − 𝑠 2 )

Note In some cases the direct substitution will be more convenient whereas in other cases the
chain rule will be more useful. Read and follow instructions carefully.

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 dz
Example Using the chain rule, find , given that z = x 2 − y 2 ; x = cost , y = sin t
dt

Solution Since the question clearly states “using the chain rule”, then that is the method to
be applied.

𝑑𝑧 𝜕𝑧 𝑑𝑥 𝜕𝑧 𝑑𝑦
= × + × = (2𝑥)(−𝑠𝑖𝑛𝑡) + (−2𝑦)(𝑐𝑜𝑠𝑡) =
𝑑𝑡 𝜕𝑥 𝑑𝑡 𝜕𝑦 𝑑𝑡

−2𝑐𝑜𝑠𝑡𝑠𝑖𝑛𝑡 − 2𝑠𝑖𝑛𝑡𝑐𝑜𝑠𝑡 = −4𝑠𝑖𝑛𝑡𝑐𝑜𝑠𝑡

*If you are feeling idle, you could check your answer using direct substitution:
𝑑𝑧
𝑧 = (𝑐𝑜𝑠𝑡)2 − (𝑠𝑖𝑛𝑡)2 ; = −2𝑠𝑖𝑛𝑡𝑐𝑜𝑠𝑡 − 2𝑐𝑜𝑠𝑡𝑠𝑖𝑛𝑡 = −4𝑠𝑖𝑛𝑡𝑐𝑜𝑠𝑡
𝑑𝑡

𝑥 = 𝑢2 − 𝑣 2
Example Consider 𝑧 = 2𝑥 2 + 3𝑦 3 ; with { .
𝑦 = 2𝑢𝑣
𝜕𝑧 𝜕𝑧
Find expressions, in terms of u and v only, for and , using:
𝜕𝑢 𝜕𝑣
a) Direct substitution
b) Chain rule

Solution

a) By direct substitution; 𝑧 = 2(𝑢2 − 𝑣 2 )2 + 3(2𝑢𝑣)3 = 2(𝑢2 − 𝑣 2 )2 + 24𝑢3 𝑣 3

𝜕𝑧 𝜕𝑧
Then, = 8𝑢(𝑢2 − 𝑣 2 ) + 72𝑢2 𝑣 3 and = −8𝑣(𝑢2 − 𝑣 2 ) + 72𝑢3 𝑣 2 .
𝜕𝑢 𝜕𝑣

b) By chain rule:
𝜕𝑧 𝜕𝑧 𝜕𝑥 𝜕𝑧 𝜕𝑦
= × + × = 4𝑥(2𝑢) + 9𝑦 2 (2𝑣) = 8𝑢(𝑢2 − 𝑣 2 ) + 18𝑣(2𝑢𝑣)2
𝜕𝑢 𝜕𝑥 𝜕𝑢 𝜕𝑦 𝜕𝑢
= 8𝑢(𝑢2 − 𝑣 2 ) + 72𝑢2 𝑣 3
And
𝜕𝑧 𝜕𝑧 𝜕𝑥 𝜕𝑧 𝜕𝑦
= × + × = 4𝑥(−2𝑣) + 9𝑦 2 (2𝑢)
𝜕𝑣 𝜕𝑥 𝜕𝑣 𝜕𝑦 𝜕𝑣
= −8𝑣(𝑢2 − 𝑣 2 ) + 18𝑢(2𝑢𝑣)2 = −8𝑣(𝑢2 − 𝑣 2 ) + 72𝑢3 𝑣 2

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 𝜕𝑧 𝜕𝑧 𝜕𝑧 𝜕𝑧
Example If 𝑧 = 𝑓(𝑥, 𝑦); 𝑥 = 𝑒 𝑢 + 𝑒 −𝑣 , 𝑦 = 𝑒 −𝑢 − 𝑒 𝑣 , prove that − =𝑥 −𝑦
𝜕𝑢 𝜕𝑣 𝜕𝑥 𝜕𝑦

Solution Direct substitution is not applicable here, as we have no function to substitute in,
hence the Chain rule will be applied. Starting with the LHS, let us derive
𝜕𝑧 𝜕𝑧
expressions for and , using the chain rule:
𝜕𝑢 𝜕𝑣

𝜕𝑧 𝜕𝑧 𝜕𝑥 𝜕𝑧 𝜕𝑦 𝜕𝑧 𝜕𝑧
= × + × = × 𝑒𝑢 + × (−𝑒 −𝑢 )
𝜕𝑢 𝜕𝑥 𝜕𝑢 𝜕𝑦 𝜕𝑢 𝜕𝑥 𝜕𝑦

𝜕𝑧 𝜕𝑧 𝜕𝑥 𝜕𝑧 𝜕𝑦 𝜕𝑧 𝜕𝑧
= × + × = × (−𝑒 −𝑣 ) + × (−𝑒 𝑣 )
𝜕𝑣 𝜕𝑥 𝜕𝑣 𝜕𝑦 𝜕𝑣 𝜕𝑥 𝜕𝑦

Substituting in the LHS gives:

𝜕𝑧 𝜕𝑧 𝜕𝑧 𝜕𝑧 𝜕𝑧 𝜕𝑧
− = × 𝑒𝑢 + × (−𝑒 −𝑢 ) − [ × (−𝑒 −𝑣 ) + × (−𝑒 𝑣 )] =
𝜕𝑢 𝜕𝑣 𝜕𝑥 𝜕𝑦 𝜕𝑥 𝜕𝑦

𝜕𝑧 𝜕𝑧 𝜕𝑧 𝜕𝑧
𝑒𝑢 − 𝑒 −𝑢 + 𝑒 −𝑣 + 𝑒𝑣
𝜕𝑥 𝜕𝑦 𝜕𝑥 𝜕𝑦

Grouping like terms:

𝜕𝑧 𝜕𝑧 𝜕𝑧 𝜕𝑧 𝜕𝑧 𝜕𝑧
(𝑒 𝑢 + 𝑒 −𝑣 ) + (𝑒 𝑣 − 𝑒 −𝑢 ) = (𝑒 𝑢 + 𝑒 −𝑣 ) − (𝑒 −𝑢 − 𝑒 𝑣 ) = 𝑥 −𝑦
𝜕𝑥 𝜕𝑦 𝜕𝑥 𝜕𝑦 𝜕𝑥 𝜕𝑦

𝑑𝑧
Example Using the chain rule, find , for the following, in terms of t:
𝑑𝑡

a) 𝑧 = 5𝑥 − 3𝑦 2 ; 𝑥 = 2𝑐𝑜𝑠 𝑡 , 𝑦 = 3 𝑠𝑖𝑛 𝑡
b) 𝑧 = 𝑠𝑖𝑛(𝑥𝑦) ; 𝑥 = 𝑒 −𝑡 , 𝑦 = 3𝑡 2

Solution
𝑑𝑧
a) = (5)(−2𝑠𝑖𝑛𝑡) + (−6𝑦)(3𝑐𝑜𝑠𝑡) = −10𝑠𝑖𝑛𝑡 − 18𝑦𝑐𝑜𝑠𝑡 =
𝑑𝑡

−10𝑠𝑖𝑛𝑡 − 54𝑠𝑖𝑛𝑡 𝑐𝑜𝑠𝑡

𝑑𝑧
b) = (𝑦𝑐𝑜𝑠(𝑥𝑦))(−𝑒 −𝑡 ) + (𝑥𝑐𝑜𝑠(𝑥𝑦))(6𝑡) =
𝑑𝑡

−3𝑡 2 𝑒 −𝑡 𝑐𝑜𝑠(3𝑡 2 𝑒 −𝑡 ) + 6𝑡𝑒−𝑡 𝑐𝑜𝑠(3𝑡 2 𝑒 −𝑡 )

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 𝜕𝑧 𝜕𝑧
Example Using the chain rule, find and , for the following:
𝜕𝑟 𝜕𝑠

3
a) 𝑧 = 3𝑥 + 𝑦 3 + 2𝑡 2 ; 𝑥 = 𝑟𝑒 −𝑠 , 𝑦 = 𝑠𝑒 𝑟 , 𝑡 = 3𝑟 2 ( √𝑠)
𝑟
b) 𝑧 = 5𝑥 − 2𝑦; 𝑥= , 𝑦 = 2𝑟 + 𝑙𝑛 𝑠
𝑠2

Solution

a) In this example 𝑧 = 𝑓(𝑥, 𝑦, 𝑡) and 𝑥, 𝑦 𝑎𝑛𝑑 𝑡 are each a function of 𝑠 𝑎𝑛𝑑 𝑟.

𝜕𝑧 𝜕𝑧 𝜕𝑥 𝜕𝑧 𝜕𝑦 𝜕𝑧 𝜕𝑡 𝜕𝑧 𝜕𝑧 𝜕𝑥 𝜕𝑧 𝜕𝑦 𝜕𝑧 𝜕𝑡
= × + × + × and = × + × + ×
𝜕𝑟 𝜕𝑥 𝜕𝑟 𝜕𝑦 𝜕𝑟 𝜕𝑡 𝜕𝑟 𝜕𝑠 𝜕𝑥 𝜕𝑠 𝜕𝑦 𝜕𝑠 𝜕𝑡 𝜕𝑠

𝜕𝑧 3
= (3)(𝑒 −𝑠 ) + (3𝑦 2 )(𝑠𝑒 𝑟 ) + (4𝑡)(6𝑟( √𝑠))
𝜕𝑟
3 3
= 3𝑒 −𝑠 + 3𝑠 2 𝑒 2𝑟 𝑠𝑒 𝑟 + 12𝑟 2 ( √𝑠) (6𝑟( √𝑠))
3 2
= 3𝑒 −𝑠 + 3𝑠 3 𝑒 3𝑟 + 72𝑟 3 ( √𝑠)

𝜕𝑧 2
= (3)(−𝑟𝑒 −𝑠 ) + (3𝑦 2 )(𝑒 𝑟 ) + (4𝑡) (𝑟 2 𝑠 − 3 )
𝜕𝑠
2
= −3𝑟𝑒 −𝑠 + 3𝑒 𝑟 (𝑠𝑒 𝑟 )2 + 12𝑟 2 ( √𝑠)𝑟 2 𝑠 − 3
3

1
−
−𝑠 2 3𝑟 3𝑠 3
= −3𝑟𝑒 + 3𝑠 𝑒 + 12𝑟

𝜕𝑧 𝜕𝑧 𝜕𝑥 𝜕𝑧 𝜕𝑦 5
b) = × + × = (5)(𝑠 −2 ) + (−2)(2) = −4
𝜕𝑟 𝜕𝑥 𝜕𝑟 𝜕𝑦 𝜕𝑟 𝑠2

𝜕𝑧 𝜕𝑧 𝜕𝑥 𝜕𝑧 𝜕𝑦 −3
1 −3
2𝑠 2
= × + × = (5)(−𝑟𝑠 ) + (−2) ( ) = −5𝑟𝑠 −
𝜕𝑠 𝜕𝑥 𝜕𝑠 𝜕𝑦 𝜕𝑠 𝑥 𝑟

Prepared by T-A Russell (tarussell@utech.edu.jm) July 8, 2020

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