This document discusses partial derivatives, which are used to describe the rate of change of functions with multiple variables. It defines: 1) Partial derivatives as the rate of change of the dependent variable with respect to one independent variable, while holding other variables constant. 2) Functions of two variables have level curves where the function value is constant. Their graphs are surfaces in 3D space. 3) Higher order partial derivatives describe the rate of change of the first partial derivatives. 4) The chain rule extends differentiation to composite functions, allowing functions of variables that are themselves functions of other variables.

1. BMM 104: ENGINEERING MATHEMATICS I Page 1 of 8
CHAPTER 5: PARTIAL DERIVATIVES
Functions of n Independent Variables
Suppose D is a set of n-tuples of real numbers ( x1 , x 2 ,..., x n ) . A real valued function f
on D is a rule that assigns a unique (single) real number
w = f ( x1 , x 2 ,..., x n )
to each element in D. The set D is the function’s domain. The set of w-values taken on
by f is the function’s range. The symbol w is the dependent variable of f, and f is said to
be a function of the n independent variables x1 to x n . We also call the x j ' s the
function’s input variables and call w the function’s output variable.
Level Curve, Graph, surface of Functions of Two Variables
The set of points in the plane where a function f ( x , y ) has a constant value f ( x , y ) = c
is called a level curve of f. The set of all points ( x , y , f ( x , y ) ) in space, for ( x , y ) in
the domain of f, is called the graph of f. The graph of f is also called the surface
z = f ( x , y) .
Functions of Three Variables
The set of points ( x , y , z ) in space where a function of three independent variables has a
constant value f ( x , y , z ) = c is called a level surface of f.
Example: Attend lecture.
Partial Derivatives of a Function of Two Variables
Definition: Partial Derivative with Respect to x
The partial derivative of f ( x , y ) with respect to x at the point ( x0 , y 0 ) is
∂f f ( x0 + h , y 0 ) − f ( x0 , y 0 )
= lim ,
∂x ( x0 , y0 )
h→0 h
provided the limit exists.
Definition: Partial Derivative with Respect to y

2. BMM 104: ENGINEERING MATHEMATICS I Page 2 of 8
The partial derivative of f ( x , y ) with respect to y at the point ( x0 , y0 ) is
∂f d f ( x0 , y 0 + h ) − f ( x0 , y 0 )
= f ( x0 , y ) = lim ,
∂y ( x0 , y0 ) dy y = y0 h→0 h
provided the limit exists.
Example:
∂f ∂f
1. Find the values of and ∂ at the point ( 4 ,− ) if
5
∂x y
f ( x , y ) = x 2 + 3 xy + y − 1 .
∂ f
2. Find if f ( x , y ) = y sin xy .
∂ x
2y
3. Find f x and f y if f ( x , y ) = y + cos x .
Functions of More Than Two Variables
Example:
∂f ∂f ∂f
1. Let f ( x , y , z ) = xy 2 z 3 . Find , ∂ and at (1,− ,− ) .
2 1
∂x y ∂z
y
2. Let g ( x , y , z ) = x 2 e z . Find g x , g y and g z .
Second-Order Partial Derivatives
When we differentiate a function f ( x , y ) twice, we produce its second-order
derivatives.
These derivatives are usually denoted by
∂2 f f xx
“ d squared fdx squared “ or “f sub xx “
∂x 2
∂2 f
“ d squared fdy squared “ or f yy “f sub yy “
∂ 2
y
∂2 f f xx
“ d squared fdx squared “ or “f sub xx “
∂x 2
∂ f
2
“ d squared fdxdy squared “ or f yx “f sub yx “
∂∂
x y

3. BMM 104: ENGINEERING MATHEMATICS I Page 3 of 8
∂ f
2
“ d squared fdydx squared “ or f xy “f sub xy “
∂∂
y x
The defining equations are
∂2 f ∂  ∂f  ∂2 f ∂ ∂ 
f
=  , = 
∂ 
∂x 2
∂x  ∂x  ∂∂
x y ∂  y
x 
and so on. Notice the order in which the derivatives are taken:
∂ f
2
Differentiate first with respect to y, then with respect to x.
∂∂
x y
f yx = ( f y ) x Means the same thing.
Example:
∂2 f ∂ f
2
∂ f ∂ f
2 2
1. Let f ( x , y ) = x 3 y 2 − x 4 y 6 . Find , , and .
∂x 2 ∂ ∂ ∂
y x y2 ∂∂x y
∂2 f ∂ f
2
∂2 f ∂ f
2
2. If f ( x , y ) = x cos y + ye x , find , , and .
∂x 2 ∂ ∂ ∂
y x y2 ∂∂x y
The Chain Rule
Chain Rule for Functions of Two Independent Variables
If w = f ( x , y ) has continuous partial derivatives f x and f y and if x = x( t ) , y = y ( t )
are
differentiable functions of t, then the compose w = f ( x( t ) , y ( t ) ) is a differentiable
function of t and
df
= f x ( x ( t ) , y ( t ) ) • x ' ( t ) + f y ( x( t ) , y ( t ) ) • y ' ( t ) ,
dt
or
dw ∂f dx ∂f dy
= + .
dt ∂x dt ∂y dt
Example:
Use the chain rule to find the derivative of w = xy , with respect to t along the path

4. BMM 104: ENGINEERING MATHEMATICS I Page 4 of 8
π
x = cos t , y = sin t . What is the derivative’s value at t = ?
2
Chain Rule for Functions of Three Independent Variables
If w = f ( x , y , z ) is differentiable and x, y and z are differentiable functions of t, then w
is
a differentiable function of t and
dw ∂f dx ∂f dy ∂f dz
= + + .
dt ∂x dt ∂y dt ∂z dt
Example:
dw
Find if w = xy + z , x = cos t , y = sin t , z =t.
dt
Chain Rule for Two Independent Variables and Three Intermediate Variables
Suppose that w = f ( x , y , z ) , x = g ( r , s ) , y = h( r , s ) , and z = k ( r , s ) . If all four
functions
are differentiable, then w has partial derivatives with respect to r and s, given by the
formulas
∂w ∂ ∂
w x ∂ ∂w y ∂ ∂
w z
= + +
∂r ∂ ∂
x r ∂ ∂
y r ∂ ∂
z r
∂w ∂ ∂
w x ∂ ∂w y ∂ ∂
w z
= + +
∂s ∂ ∂
x s ∂ ∂
y s ∂ ∂
z s
Example:
∂w ∂w
Express and in terms of r and s is
∂r ∂s
r
w = x +2y + z2, x = , y = r 2 + ln s , z = 2 r .
s
If w = f ( x , y ) , x = g ( r , s ) , and y = h( r , s ) , then
∂w ∂ ∂
w x ∂ ∂
w y ∂w ∂ ∂
w x ∂ ∂
w y
= + and = +
∂r ∂ ∂
x r ∂ ∂
y r ∂s ∂ ∂
x s ∂ ∂
y s
Example:
∂w ∂w
Express and in terms of r and s if
∂r ∂s

5. BMM 104: ENGINEERING MATHEMATICS I Page 5 of 8
w = x2 + y2 , x= r− s, y =r+s.
If w = f ( x ) and x = g ( r , s ) , then
∂w dw ∂x ∂w dw ∂x
= and = .
∂r dx ∂r ∂s dx ∂s
PROBLEM SET: CHAPTER 5
1. Sketch and name the surfaces
(a) f ( x, y , z ) = x 2 + y 2 + z 2 (e) f ( x, y, z ) = x 2 + y 2
(b) f ( x , y , z ) = ln( x 2 + y 2 + z 2 ) (f) f ( x, y, z ) = y 2 + z 2
(c) f ( x, y , z ) = x + z (g) f ( x, y, z ) = z − x 2 − y 2
x2 y2 z2
(d) f ( x, y, z ) = z (h) f ( x, y , z ) = + +
25 16 9
∂f ∂f
2. Find and ∂ .
∂x y
(a) f ( x , y ) = 5 xy − 7 x 2 − y 2 + 3 x − 6 y + 2
y
(b) f ( x , y ) = tan −1  
x
(c) f ( x, y) = e ( x + y +1)
(d) f ( x , y ) = e −x sin( x + y )
(e) f ( x , y ) = ln( x + y )
(f) f ( x , y ) = sin 2 ( x − 3 y )
3. Find f x , f y and f z .
(a) f ( x , y , z ) = sin −1 ( xyz )
f ( x , y , z ) = e −( x )
2
+ y 2 +z 2
(b)
(c) f ( x , y , z ) = e −xyz
(d) f ( x , y , z ) = tanh( x + 2 y + 3 z )
4. Find all the second-order partial derivatives of the following functions.
(a) f ( x , y ) = x + y + xy
(b) f ( x , y ) = sin xy
(c) f ( x , y ) = x 2 y + cos y + y sin x

6. BMM 104: ENGINEERING MATHEMATICS I Page 6 of 8
(d) f ( x , y ) = xe y + y + 1
5. Verify that w xy = w yx .
(a) w = ln( 2 x + 3 y ) (c) w = xy 2 + x 2 y 3 + x 3 y 4
(b) w = e x + x ln y + y ln x (d) w = x sin y + y sin x + xy
dw
6. In the following questions, (a) express as a function of t, both by using
dt
the Chain Rule and by expressing w in terms of t and differentiating directly with
dw
respect to t. The (b) evaluate at the given value of t.
dt
(i) w = x2 + y2 , x = cos t , y = sin t ; t=π .
x y 1
(ii) w= + , x = cos 2 t , y = sin 2 t , z= t =3.
z z t
∂z ∂z
7. In the following questions, (a) express and as a functions of u and v
∂u ∂v
both by using the Chain Rule and by expressing z directly in terms of u and v
∂z ∂z
before differentiating. Then (b) evaluate and at the given point (u , v ) .
∂u ∂v
(i) z = 4 e x ln y , x = ln( u cos v ) , y = u sin v ; ( u ,v ) =  2 , π 
 
 4
(ii)
x
z = tan −1   ,
y x = u cos v , y = u sin v ; ( u ,v ) = 1.3 , π 
 
   6
ANSWERS FOR PROBLEM SET: CHAPTER 5
∂f ∂f
2. (a) = 5 y − 14 x + 3 , = 5 x − 2 y −6
∂x ∂y
∂f y ∂f x
(b) =− 2 , = 2
∂x x + y 2 ∂y x + y2
∂f ∂f
(c) = e ( x +y +1) , = e ( x +y +1 )
∂x ∂y
∂f ∂f
(d) = −e −x sin( x + y ) + e −x cos ( x + y ) , = e −x cos ( x + y )
∂x ∂y
∂f 1 ∂f 1
(e) = , =
∂x x + y ∂y x+y
∂f ∂f
(f) = 2 sin( x − 3 y ) cos( x − 3 y ) , = −6 sin( x − 3 y ) cos( x − 3 y )
∂x ∂x

7. BMM 104: ENGINEERING MATHEMATICS I Page 7 of 8
yz xz xy
3. (a) fx = , fy = , fz =
1−x y z2 2 2
1−x y z
2 2 2
1 − x2 y2 z2
f x = −2 xe −( x ) , f = −2 ye −( x f z = −2 ze − ( x + y + z )
(b)
2
+ y 2 +z 2 2
+ y 2 +z 2 ) , 2 2 2
y
(c) f x = −yze −xyz , f y = −xze −xyz , f z = −xye − xyz
(d) f x = sec h 2 ( x + 2 y + 3 z ) , f y = 2 sec h 2 ( x + 2 y + 3 z ) ,
f z = 3 sec h 2 ( x + 2 y + 3 z )
∂f ∂f ∂2 f
=1 + x , ∂ f = 0,
2
∂2 f ∂2 f
4. (a) = 1 + y, = 0, = =1
∂x ∂y ∂x 2 ∂y 2 ∂∂
y x ∂∂
x y
∂f ∂f ∂2 f
= x cos xy , ∂ f = − y 2 sin xy ,
2
(b) = y cos xy , = −x 2 sin xy ,
∂x ∂y ∂x 2 ∂y 2
∂2 f ∂2 f
= = cos xy − xy sin xy
∂y∂x ∂x∂y
∂f ∂f
= x 2 − sin y + sin x , ∂ f = 2 y − y sin x ,
2
(c) = 2 xy + y cos x ,
∂x ∂y ∂x 2
∂ f
2
∂2 f ∂2 f
= −cos y , = = 2 x + cos x
∂y 2
∂y∂x ∂x∂y
∂f ∂f ∂2 f
= xe y + 1 , ∂ f = 0 ,
2
∂2 f ∂2 f
(d) =ey = xe y , = =e y
∂x ∂y ∂x 2 ∂y 2
∂∂
y x ∂∂
x y
∂w 2 ∂w 3 ∂2 w −6
5. (a) = , = , = , and
∂x 2 x + 3 y ∂y 2 x + 3 y ∂y∂x (2x + 3 y) 2
∂2 w −6
=
∂x∂y ( 2 x + 3 y ) 2
∂w y ∂w x ∂2 w 1 1
(b) = e x + ln y + , = + ln x , = + , and
∂x x ∂y y ∂∂
y x y x
∂2w 1 1
= +
∂x∂y y x
∂w ∂w
(c) = y 2 + 2 xy 3 + 3 x 2 y 4 , = 2 xy + 3 x 2 y 2 + 4 x 3 y 3 ,
∂x ∂y
∂2 w ∂2 w
= 2 y + 6 xy 2 + 12 x 2 y 3 , and = 2 y + 6 xy 2 + 12 x 2 y 3
∂y∂x ∂x∂y

8. BMM 104: ENGINEERING MATHEMATICS I Page 8 of 8
∂w ∂w
(d) = sin y + y cos x + y , = x cos y + sin x + x ,
∂x ∂y
∂2 w ∂2 w
= cos y + cos x + 1, and = cos y + cos x + 1
∂y∂x ∂x∂y
dw
6. (i) (a) =0 (b) 0
dt
dw
(ii) (a) =1 (b) 1
dt
∂z
7. (i) (a) = ( 4 cos v ) ln( u sin v ) + 4 cos v
∂u
∂z 4u cos 2 v
= ( − 4u sin v ) ln( u sin v ) +
∂v sin v
∂z
(b) = 2 ( ln 2 + 2 )
∂u
∂z
= −2 2 ln 2 + 4 2
∂v
∂z
(ii) (a) =0
∂u
∂z
= −1
∂v
∂z
(b) =0
∂u
∂z
= −1
∂v

9. BMM 104: ENGINEERING MATHEMATICS I Page 8 of 8
∂w ∂w
(d) = sin y + y cos x + y , = x cos y + sin x + x ,
∂x ∂y
∂2 w ∂2 w
= cos y + cos x + 1, and = cos y + cos x + 1
∂y∂x ∂x∂y
dw
6. (i) (a) =0 (b) 0
dt
dw
(ii) (a) =1 (b) 1
dt
∂z
7. (i) (a) = ( 4 cos v ) ln( u sin v ) + 4 cos v
∂u
∂z 4u cos 2 v
= ( − 4u sin v ) ln( u sin v ) +
∂v sin v
∂z
(b) = 2 ( ln 2 + 2 )
∂u
∂z
= −2 2 ln 2 + 4 2
∂v
∂z
(ii) (a) =0
∂u
∂z
= −1
∂v
∂z
(b) =0
∂u
∂z
= −1
∂v

10. BMM 104: ENGINEERING MATHEMATICS I Page 8 of 8
∂w ∂w
(d) = sin y + y cos x + y , = x cos y + sin x + x ,
∂x ∂y
∂2 w ∂2 w
= cos y + cos x + 1, and = cos y + cos x + 1
∂y∂x ∂x∂y
dw
6. (i) (a) =0 (b) 0
dt
dw
(ii) (a) =1 (b) 1
dt
∂z
7. (i) (a) = ( 4 cos v ) ln( u sin v ) + 4 cos v
∂u
∂z 4u cos 2 v
= ( − 4u sin v ) ln( u sin v ) +
∂v sin v
∂z
(b) = 2 ( ln 2 + 2 )
∂u
∂z
= −2 2 ln 2 + 4 2
∂v
∂z
(ii) (a) =0
∂u
∂z
= −1
∂v
∂z
(b) =0
∂u
∂z
= −1
∂v

11. BMM 104: ENGINEERING MATHEMATICS I Page 8 of 8
∂w ∂w
(d) = sin y + y cos x + y , = x cos y + sin x + x ,
∂x ∂y
∂2 w ∂2 w
= cos y + cos x + 1, and = cos y + cos x + 1
∂y∂x ∂x∂y
dw
6. (i) (a) =0 (b) 0
dt
dw
(ii) (a) =1 (b) 1
dt
∂z
7. (i) (a) = ( 4 cos v ) ln( u sin v ) + 4 cos v
∂u
∂z 4u cos 2 v
= ( − 4u sin v ) ln( u sin v ) +
∂v sin v
∂z
(b) = 2 ( ln 2 + 2 )
∂u
∂z
= −2 2 ln 2 + 4 2
∂v
∂z
(ii) (a) =0
∂u
∂z
= −1
∂v
∂z
(b) =0
∂u
∂z
= −1
∂v