UNIT 3
MULTIVARIABLE CALCULUS (DIFFERENTIATION)
Function of two variables- R denotes the set of real numbers. Let’s D is a collection of pairs of real numbers (x, y), which means D is a subset of R × R.
Then a real valued function of two variables of f is a rule that assign to each point (x,y) in D a unique real number denoted by f(x,y).
The set D is called the domain of f.
The set [ f(x,y) : (x,y) belongs to D ], which is the set of values the function f takes, is called range of f.
Here we generally use the letter z to denote the value that a function of two variables takes,
Then we will have,
z = f(x, y)
Here we will call that z is the dependent variable and x and y are independent variables.
Example-1: The area of a rectangular figure whose length is l and breadth is b, given by l × b.
Here independent variables are l and b, but dependent variable is area.
Example-2: the volume of a cylinder is given by πr²h, where r is the radius and h is the height of the cylinder.
In which radius and height are independent variables and volume is dependent.
Example-3: the volume of cuboid is given by l × b× h. Where l, b and h are the length, breadth and height respectively.
L, b and h are independent variable and volume of cuboid is dependent variable.
Limits-
The function f(x,y) is said to tend to limit ‘l’ , as x →a and y→b Iff the limit is dependent on point (x,y) as x →a and y→b
We can write this as,
Example-1: evaluate the 
Sol. We can simply find the solution as follows,
Example-2: evaluate 
Sol.
-6.
Example-3: evaluate 
Sol.
Conitinuity –
At point (a,b) , a function f(x,y) is said to be continuous if,
Working rule for continuity-
Step-1: f(a,b) should be well defined.
Step-2: 
should exist.
Step-3: 
Example-1: Test the continuity of the following function-
Sol. (1) the function is well defined at (0,0)
(2) check for the second step,
That means the limit exists at (0,0)
Now check step-3:
So that the function is continuous at origin.
Example-2: check for the continuity of the following function at origin,
Sol. (1) Here the function is well defined at (0,0)
(2) Check for second step-
Limit f is not unique for different values of m
So that the limit does not exists.
Therefore the function is not continuous at origin.
Steps to check for existence of limit-
Step-1: find the value of f(x,y) along x →a and y→b
Step-2: find the value of f(x,y) along x →b and y→a
Note- if the values in step -1 and step-2 are same then we can say that the limits exist otherwise not.
Step-3: if a →0 and b→0 then find the limit along y =mx, if the value does not contain m then limit exist, If it contains m then the limit does not exist.
Note-1- put x = 0 and y = 0 in f, then find f1
2 - Put y = 0 and x = 0 In f then find x2
If f1 and f2 are equal then limit exist otherwise not.
3- put y = mx then find f3
If f1 = f2 ≠f3, limit does not exist.
4- put y = mx² and find f4,
If f1 = f2 = f3 ≠ f4, limit does not exist
If f1 = f2 = f3 = f4, limit exist.
Example-1: Evaluate 
Sol . 1. 
2. 
Here f1 = f2
3. Now put y = mx, we get
Here f1 = f2 = f3
Now put y = mx²
4. 
Therefore,
F1 = f2 = f3 =f4
We can say that the limit exists with 0.
Example-2: evaluate the following-
Sol. First we will calculate f1 –
Here we see that f1 = 0
Now find f2,
Here , f1 = f2
Therefore the limit exists with value 0.
First order partial differentiation-
Let f(x, y) be a function of two variables. Then the partial derivative of this function with respect to x can be written as 
and defined as follows:
= 
Now the partial derivative of f with respect to f can be written as 
and defined as follows:
= 
Note: a. While calculating partial derivatives treat all independent variables, other than the variable with respect to which we are differentiating, as constant.
b. We apply all differentiation rules.
Higher order partial differentiation-
Let f(x , y) be a function of two variables. Then its second-order partial derivatives, third order partial derivatives and so on are referred as higher order partial derivatives.
These are second order four partial derivatives:
(a) 
= 
(b) 
= 
(c) 
= 
(d) 
= 
b and c are known as mixed partial derivatives.
Similarly we can find the other higher order derivatives.
Example-1: - Calculate 
and 
for the following function
f(x , y) = 3x³-5y²+2xy-8x+4y-20
Sol. To calculate 
treat the variable y as a constant, then differentiate f(x,y) with respect to x by using differentiation rules,
= 
[3x³-5y²+2xy-8x+4y-20]
= 
3x³] - 
5y²] + 
[2xy] -
8x] +
4y] - 
20]
= 9x² - 0 + 2y – 8 + 0 – 0
= 9x² + 2y – 8
Similarly partial derivative of f(x,y) with respect to y is:
= 
[3x³-5y²+2xy-8x+4y-20]
= 
3x³] - 
5y²] + 
[2xy] -
8x] +
4y] - 
20]
= 0 – 10y + 2x – 0 + 4 – 0
= 2x – 10y +4.
Example-2: Calculate 
and 
for the following function
f( x, y) = sin(y²x + 5x – 8)
Sol. To calculate 
treat the variable y as a constant, then differentiate f(x,y) with respect to x by using differentiation rules,
[sin(y²x + 5x – 8)]
= cos(y²x + 5x – 8)
(y²x + 5x – 8)
= (y² + 50) cos(y²x + 5x – 8)
Similarly partial derivative of f(x,y) with respect to y is,
[sin(y²x + 5x – 8)]
= cos(y²x + 5x – 8)
(y²x + 5x – 8)
= 2xy cos(y²x + 5x – 8)
Example-3: Obtain all the second order partial derivative of the function:
f( x, y) = ( x³y² - xy⁵)
Sol. 
3x²y² - y⁵, 
2x³y – 5xy⁴,
= 
= 6xy²
= 
2x³ - 20xy³
= 
= 6x²y – 5y⁴
= 
= 6x²y - 5y⁴
Example-4: Find
Sol. First we will differentiate partially with repsect to r,
Now differentiate partially with respect to θ, we get
Example-5: if,
Then find.
Sol-
Example-6: if 
, then show that- 
Sol. Here we have,
u = 
…………………..(1)
Now partially differentiate eq.(1) w.r to x and y , we get
= 
Or
………………..(2)
And now,
= 
………………….(3)
Adding eq. (1) and (3) , we get
= 0
Hence proved.
When we measure the rate of change of the dependent variable owing to any change in a variable on which it depends, when none of the variable is assumed to be constant.
Let the function, u = f( x, y), such that x = g(t) , y = h(t)
ᵡ Then we can write,
= 
= 
This is the total derivative of u with respect to t.
Change of variable-
If w = f (x, y) has continuous partial variables fx and fy and if x = x (t), y = y (t) are
Differentiable functions of t, then the composite function w = f (x (t), y (t)) is a
Differentiable function of t.
In this case, we get,
fx (x (t), y (t)) x’ (t) + fy (x(t), y (t)) y’ (t).
Example-:1 let q = 4x + 3y and x = t³ + t² + 1 , y = t³ - t² - t
Then find 
.
Sol. : 
. = 
Where, f1 = 
, f2 = 
In this example f1 = 4 , f2 = 3
Also, 
3t² + 2t , 
4(3t² + 2t) + 3(
= 21t² + 2t – 3
Example-2: Find 
if u = x³y⁴ where x = t³ and y = t².
Sol. As we know that by definition, 
= 
3x²y⁴3t² + 4x³y³2t = 17t¹⁶.
Example-3: if w = x² + y – z + sint and x + y = t, find
(a) 
y,z
(b) 
t, z
Sol. With x, y, z independent, we have
t = x + y, w = x² + y - z + sin (x + y).
Therefore,
y,z = 2x + cos(x+y)
(x+y)
= 2x + cos (x + y)
With x, t, z independent, we have
Y = t-x, w= x² + (t-x) + sin t
Thus 
t, z = 2x - 1
Example-4: If u = u( y – z , z - x , x – y) then prove that 
= 0
Sol. Let,
Then,
By adding all these equations we get,
= 0 hence proved.
Example-5: if φ( cx – az , cy – bz) = 0 then show that ap + bq = c
Where p = 
q = 
Sol. We have,
φ( cx – az , cy – bz) = 0
φ( r , s) = 0
Where,
We know that,
Again we do,
By adding the two results, we get
Example-6: If z is the function of x and y , and x = 
, y = 
, then prove that,
Sol. Here , it is given that, z is the function of x and y & x , y are the functions of u and v.
So that,
……………….(1)
And,
………………..(2)
Also there is,
x = 
and y = 
,
Now,
, 
, 
, 
From equation(1) , we get
……………….(3)
And from eq. (2) , we get
…………..(4)
Subtracting eq. (4) from (3), we get
= 
) 
– (
= x
Hence proved.
Differentiability-
Let f(x) be a single valued function of the variable x, then,
f’(x) = 
Provided that the limit exists and is dependent of the path along which 
Example: suppose that the function,
f(z) = 4x + y + i( -x + 4y)
Discuss df/dz.
Sol. Here,
Example-2: if
Then df/dz , z = 0.
Sol.
A composite function is a composition / combination of the functions. In this value of one function depends on the value of another function. A composite function is created when one function is put in another.
Let 
i.e 
To differentiate composite function chain rule is used:
Chain rule:
- If
where x,y,z are all the function of t then
2. If 
be an implicit relation between x and y .
Differentiating with respect to x we get
We get 
Example 1 : If 
where 
then find the value of 
?
Given 
Where 
By chain rule
Now substituting the value of x ,y,z we get
-6
8
Example2 :If 
then calculate 
Given 
By Chain Rule
Putting the value of u = 
Again partially differentiating z with respect to y
By Chain Rule
by substituting value 
Example 3 :If 
.
Show that 
Given 
Partially differentiating u with respect to x and using chain rule
………(i)
Partially differentiating z with respect to y and using chain rule
= 
………..(ii)
Partially differentiating z with respect to t and using chain rule
Using (i) and (ii) we get
Hence 
Example4 : If 
where the relation is 
.
Find the value of 
Let the given relation is denoted by 
We know that 
Differentiating u with respect to x and using chain rule
Example5 : If 
and the relation is 
. Find 
Given relation can be rewrite as
.
We know that
Differentiating u with respect to x and using chain rule
Implicit differentiation-
Let f(x,y) = 0
Where y = ∅(x)
By the chain rule , with x = x and y = ∅(x), we get
Here we assume that y is a differentiable funtion of x.
Example-1: if ∅ is a differentiable function such that y = ∅(x) satisfies the equation
x³ + y³ +sin xy = 0 then find 
.
Sol. Suppose f(x,y) = x³ + y³ +sin xy
Then,
fᵡ = 3x² + y cos xy
Fy = 2y + x cos xy
So ,
Example-2:
Sol. Take partial derivative on both side w.r. t. x, treat y as constant
Example-3: if x²y³ + cos y cos z = x² cos x sin y, then find 
Sol. Differentiate partially w.r.t. x and treat y as constant,
Let us suppose we have,
z = f(x , y)
Where, x = g(u , v)
And y = h(u , v)
Then,
And
Let’s do some examples-
Example-1: If u = u(
), then prove that 
.
Sol. We are given that,
u = u (
) = u( r , s)
Where r = 
and s = 
Or we can write as-
r = 
and s = 
Differentiate them partially with respect to x , y and z. , we get
As we know that-
Then,
Adding all these results, we get
.
Example-2: If u = x³ + y³ where x = a cos t and y = b sint , then find du/dt.
Sol. Given,
u = x³ + y³
x = a cost and y = b sint
= (3x²)(-a sint) + (3y²) (b cost)
= -3a³cos²t sint + 3b³sin²t cost
If u and v are functions of the two independent variables x and y , then the determinant,
Is known as the jacobian of u and v with respect to x and y, and it can be written as,
Suppose there are three functions u , v and w of three independent variables x , y and z then,
The Jacobian can be defined as,
Important properties of the Jacobians-
Property-1-
If u and v are the functions of x and y , then
Proof- Suppose u = u(x,y) and v = v(x,y) , so that u and v are the functions of x and y,
Now,
Interchange the rows and columns of the second determinant, we get
= 
= 
………….(1)
Differentiate u = u(x,y) and v= v(x,y) partially w.r.t. u and v, we get
Putting these values in eq.(1) , we get
hence proved.
Property-2:
Suppose u and v are the functions of r and s, where r,s are the fuctions of x , y, then,
Proof: 
= 
Interchange the rows and columns in second determinant
We get,
= 
= 
= 
= 
Similarly we can prove for three variables.
Property-3
If u,v,w are the functions of three independent variables x,y,z are not independent , then,
Proof: here u,v,w are independent , then f(u,v,w) = 0 ……………….(1)
Differentiate (1), w.r.t. x, y and z , we get
…………(2)
………………(3)
………………..(4)
Eliminate 
from 2,3,4 , we get
Interchanging rows and columns, we get
= 0
So that,
Example-1: If x = r sin
, y = r sin
, z = r cos
, then show that
sin
also find 
Sol. We know that,
= 
= 
= 
(on solving the determinant)
= 
Now using first property of Jacobians, we get
Example-2: If u = x + y + z , uv = y + z , uvw = z , find 
Sol. Here we have,
x = u – uv = u(1-v)
y = uv – uvw = uv( 1- w)
And z = uvw
So that,
=
Apply 
=
Now we get,
= u²v(1-w) + u²vw
= u²v
Example-3: If u = xyz , v = x² + y² + z² and w = x + y + z, then find J = 
Sol. Here u ,v and w are explicitly given , so that first we calculate
J’ = 
J’ = 
= 
= yz(2y-2z) – zx(2x – 2z) + xy (2x – 2y) = 2[yz(y-z)-zx(x-z)+xy(x-y)]
= 2[x²y - x²z - xy² + xz² + y²z - yz²]
= 2[x²(y-z) - x(y² - z²) + yz (y – z)]
= 2(y – z)(z – x)(y – x)
= -2(x – y)(y – z)(z – x)
By the property,
JJ’ = 1
J = 
When we differentiate a function depend on more than one independent variable, we differentiate it with respect to one variable keeping other as constant.
A second order partial derivative means differentiating twice
In general 
are also function of x and y and so these can be further partially differentiated with respect to x and y.
In general 
Notation:
Generalization: If 
Then the partial derivative of z with respect to 
is obtained by differentiating z with respect to 
treating all the other variables as constant and is denoted by
Example1: If 
. Then prove that
Given 
Partially differentiating z with respect to x keeping y as constant
Again partially differentiating given z with respect to y keeping x as constant
On b.eq(i) +a.eq(ii) we get
Hence proved.
Example2 : If 
Show that 
Given 
Partially differentiating z with respect to x keeping y as constant
Again partially differentiating z with respect to x keeping y as constant
Partially differentiating z with respect to y keeping x as constant
Again partially differentiating z with respect to y keeping x as constant
From eq(i) and eq(ii) we conclude that
Example3 : Find the value of n so that the equation
Satisfies the relation 
Given 
Partially differentiating V with respect to r keeping 
as constant
Again partially differentiating given V with respect to 
keeping r as constant
Now, we are taking the given relation
Substituting values using eq(i) and eq(ii)
On solving we get 
Example 4 : If 
then show that when 
Given 
Taking log on both side we get
Partially differentiating with respect to x we get
…..(i)
Similarly partially differentiating with respect y we get
……(ii)
LHS : 
Substituting value from (ii)
Again substituting value from (i) we get
]
When 
=RHS
Hence proved
Example5 :If 
Then show that 
Given 
Partially differentiating u with respect to x keeping y and z as constant
Similarly partially differentiating u with respect to y keeping x and z as constant
…….(ii)
Similarly partially differentiating u with respect to z keeping x and y as constant
…….(iii)
LHS: 
Hence proved
Taylor’s Theorem-
If f(x + h) is a function of h which can be expanded in the ascending powers of h and is differentiable by any number of times with respect to h, then-
+ …….+ 
+ ……..
Which is called Taylor’s theorem.
If we put x = a, we get-
+ …….+ 
+ …….. (1)
Maclaurin’s Theorem-
If we put a = 0 and h = x then equation(1) becomes-
+ …….
Which is called Maclaurin’s theorem.
Note – if we put h = x - a then there will be the expansion of F(x) in powers of (x – a)
We get-
+ …….
Example-1: Express the polynomial 
in powers of (x-2).
Sol. Here we have,
f(x) = 
Differentiating the function w.r.t.x-
f’(x) = 
f’’(x) = 12x + 14
f’’’(x) = 12
f’’’’(x)=0
Now using Taylor’s theorem-
+ ……. (1)
Here we have, a = 2,
Put x = 2 in the derivatives of f(x), we get-
f(2) = 
f’(2) = 
f’’(2) = 12(2)+14 = 38
f’’’(2) = 12 and f’’’’(2) = 0
Now put a = 2 and substitute the above values in equation(1), we get-
Taylor’s theorem for functions of two variables-
Suppose f(x , y) be a function of two independent variables x and y. Then,
+ ……………
Maclaurin’s series is the special case of Taylor’s series-
When we put a = 0 and b = 0 (about origin) in Taylor’s series, we get-
+ ……………
Example-2: Expand f(x , y) = 
in powers of x and y about origin.
Sol. Here we have the function-
f(x , y) = 
Here , a = 0 and b = 0 then
f(0 , 0) = 
Now we will find partial derivatives of the function-
Now using Taylor’s theorem-
+………
Suppose h = x and k = y , we get
+…….
= 
+……….
Example-3: Find the Taylor’s expansion of 
about (1 , 1) up to second degree term.
Sol. We have,
At (1 , 1)
Now by using Taylor’s theorem-
……
Suppose 1 + h = x then h = x – 1
1 + k = y then k = y - 1
……
= 
……..
As we know that the value of a function at maximum point is called maximum value of a function. Similarly the value of a function at minimum point is called minimum value of a function.
The maxima and minima of a function is an extreme biggest and extreme smallest point of a function in a given range (interval) or entire region. Pierre de Fermat was the first mathematician to discover general method for calculating maxima and minima of a function. The maxima and minima are complement of each other.
Maxima and Minima of a function of one variables
If f(x) is a single valued function defined in a region R then
Maxima is a maximum point 
if and only if 
Minima is a minimum point 
if and only if 
Maxima and Minima of a function of two independent variables
Let 
be a defined function of two independent variables.
Then the point 
is said to be a maximum point of 
if
Or 
= 
For all positive and negative values of h and k.
Similarly the point 
is said to be a minimum point of 
if
Or 
= 
For all positive and negative values of h and k.
Saddle point: Critical points of a function of two variables are those points at which both partial derivatives of the function are zero. A critical point of a function of a single variable is either a local maximum, a local minimum, or neither. With functions of two variables there is a fourth possibility - a saddle point.
A point is a saddle point of a function of two variables if
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At the point.
Stationary Value
The value 
is said to be a stationary value of 
if
i.e. the function is a stationary at (a , b).
Rule to find the maximum and minimum values of 
- Calculate
. - Form and solve
, we get the value of x and y let it be pairs of values 
- Calculate the following values :
4. (a) If 
(b) If 
(c) If 
(d) If 
Example1 Find out the maxima and minima of the function
Given 
…(i)
Partially differentiating (i) with respect to x we get
….(ii)
Partially differentiating (i) with respect to y we get
….(iii)
Now, form the equations 
Using (ii) and (iii) we get
using above two equations
Squaring both side we get
Or 
This show that 
Also we get 
Thus we get the pair of value as 
Now, we calculate
Putting above values in
At point (0,0) we get
So, the point (0,0) is a saddle point.
At point 
we get
So the point 
is the minimum point where 
In case 
So the point 
is the maximum point where 
Example2 Find the maximum and minimum point of the function
Partially differentiating given equation with respect to and x and y then equate them to zero
On solving above we get 
Also 
Thus we get the pair of values (0,0), (
,0) and (0,
Now, we calculate
At the point (0,0)
So function has saddle point at (0,0).
At the point (
So the function has maxima at this point (
.
At the point (0,
So the function has minima at this point (0,
.
At the point (
So the function has an saddle point at (
Example3 Find the maximum and minimum value of 
Let 
Partially differentiating given function with respect to x and y and equate it to zero
..(i)
..(ii)
On solving (i) and (ii) we get
Thus pair of values are 
Now, we calculate
At the point (0,0)
So further investigation is required
On the x axis y = 0 , f(x,0)=0
On the line y=x, 
At the point 
So that the given function has maximum value at 
Therefore maximum value of given function
At the point 
So that the given function has minimum value at 
Therefore minimum value of the given function
Let 
be a function of x, y, z which to be discussed for stationary value.
Let 
be a relation in x, y, z
for stationary values we have,
i.e. 
… (1)
Also from 
we have
… (2)
Let ‘
’ be undetermined multiplier then multiplying equation (2) by 
and adding in equation (1) we get,
… (3)
… (4)
… (5)
Solving equation (3), (4) (5) & we get values of x, y, z and 
.
- Decampere a positive number ‘a’ in to three parts, so their product is maximum
Solution:
Let x, y, z be the three parts of ‘a’ then we get.
… (1)
Here we have to maximize the product
i.e. 
By Lagrange’s undetermined multiplier, we get,
… (2)
… (3)
… (4)
i.e.
… (2)’
… (3)’
… (4)
And 
From (1)
Thus 
.
Hence their maximum product is 
.
2. Find the point on plane 
nearest to the point (1, 1, 1) using Lagrange’s method of multipliers.
Solution:
Let 
be the point on sphere 
which is nearest to the point 
. Then shortest distance.
Let 
Under the condition 
… (1)
By method of Lagrange’s undetermined multipliers we have
… (2)
… (3)
i.e. 
&
… (4)
From (2) we get 
From (3) we get 
From (4) we get 
Equation (1) becomes
i.e. 
y = 2
