UNIT–3
Partial Differentiation
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)
= 2xycos(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.
Homogeneous function - A function f(x,y) is said to be homogeneous of degree n if,
f(kx, ky) = kⁿf(x, y)
Here, the power of k is called the degree of homogeneity.
Or
A function f(x,y) is said to be a homogenous function in which the power of each term is the same.
Example:
1. The function-
Is a homogeneous function of order 3
Statement – if u = f(x, y) be a homogeneous function in x and y of degree n , then
x
+ y 
= nu
Proof:Here u is a homogeneous function of degree n,
u = xⁿ f(y/x) ----------------(1)
Partially differentiate equation (1) with respect to x,
= n
f(y/x) + xⁿ f’(y/x).(
)
Now multiplying by x on both sides, we get
x
= n
f(y/x) + xⁿ f’(y/x).(
) ---------- (2)
Again partially differentiate equation (1) with respect to y,
= xⁿ f’(y/x).
Now multiplying by y on both sides,
y 
= xⁿ f’(y/x).
---------------(3)
By adding equation (2) and (3),
x
y 
= n
f(y/x) + + xⁿ f’(y/x).(
) + xⁿ f’(y/x).
x
y 
= n
f(y/x)
Here u = f( x, y) is homogeneous function, then - u = 
f(y/x)
Put the value of u in equation (4),
x
y 
= nu
Which is the Euler’s theorem.
Let’s understand Eulers’s theorem by some examples:
Example1-If u = x²(y-x) + y²(x-y), then show that 
-2 (x – y)²
Solution - here, u = x²(y-x) + y²(x-y)
u = x²y - x³ + xy² - y³,
Now differentiate u partially with respect to x and y respectively,
= 2xy – 3x² + y² --------- (1)
= x² + 2xy – 3y² ---------- (2)
Now adding equation (1) and (2), we get
= -2x² - 2y² + 4xy
= -2 (x² + y² - 2xy)
= -2 (x – y)²
Example2- If u = xy + sin(xy), show that 
= 
.
Solution – u = xy + sin(xy)
= y+ ycos(xy)
= x+ xcos(xy)
x (- sin(xy).(y)) + cos(xy)
= 1 – xysin(xy) + cos(xy) -------------- (1)
1 + cos(xy) + y(-sin(xy) x)
= 1 – xysin(xy) + cos(xy) -----------------(2)
From equation (1) and (2),
= 
Example-3: If u(x,y,z) = log( tan x + tan y + tan z) , then prove that ,
Sol. Here we have,
u(x,y,z) = log( tan x + tan y + tan z) ………………..(1)
Diff. Eq.(1) w.r.t. x , partially , we get
……………..(2)
Diff. Eq.(1) w.r.t. y , partially , we get
………………(3)
Diff. Eq.(1) w.r.t. z , partially , we get
……………………(4)
Now multiply eq. 2 , 3 , 4 by sin 2x , sin 2y , sin 2z respectively and adding , in order to get the final result,
We get,
= 
So that,
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 fxand fyand 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 + sintand 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.
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 
Example1 : 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 cosxy
Fy = 2y + x cosxy
So ,
Example-2:
Sol. Take partial derivative on both side w.r. t. x , treat y as constant
Example-3: if x²y³ + cos y cosz = x²cos x sin y, then find 
Sol. Differentiate partially w.r.t. x and treat y as constant,
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 propert 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 = 
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
……
= 
……..
The real-valued functions 
of n variables 
….. 
defined on an open subset D of 
are said to be functionally dependent in D. If there exists a real valued function of F of n variables, such that-
1. One of 
, 1
, is different from zero at every point of the range of the mapping from 
which is defined by (
….. 
….. 
)
Where,
….. 
, I = 1,2,3……n
2. F(
Theorem: If u = f(x , y) and v = g(x , y) are differentiable on an open subset D of 
and are functionally dependent on D , then
Proof: from the abpvedefinition , there exists a real valued function F(u,v) such that,
F(u,v) = F(f(x,y) , g(x,y)) = 0 ………………….(1)
For all x and y which belongs to D and
Either 
at every point (u,v) of the range of the mapping
(x,y) → (f(x,y) , g(x,y))
Differentiate eq.1 partially with respect to x and y using chain rule,
And
Since for any point (x,y) which belongs to D either 
Above two equations implies that,
Example: Prove that the following functions are functionally dependent.
f(x , y) = log x – log y and g(x , y) = 
Sol. We have,
f(x , y) = log x – log y …………………(1)
g(x , y) = 
…………………………(2)
Here we will find x from the second equation and substitute in (1) , which will show the relation between the two.
Write f(x,y) = f and g(x,y) = g for convenience,
From equation(1) , we get
2g.xy = 
Lets consider,
x = y [ g + 
Put this value in eq.(1), we get
f = log x – log y
= log y [ g + 
– log y
= log y + log y ( g + 
– log y
f = log (g + 
The functional relationship F(u,v) = 0 is,
F(u,v) = u - log (v + 
Now we can verify that,
F(f(x,y) , g(x,y)) = 0
The jacobian will also be zero.
Maxima and minima of function of two variables-
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 (
Lagrange’s method-
Generally to calculate the stationary value of a function with some relation by converting the given function into the least possible independent variables and then solve them. When this method fail we use Lagrange’s method.
This method is used to calculate the stationary value of a function of several variables which are all not independent but are connected by some relation.
Let 
be the function in the variable x, y and z which is connected by the relation 
Rule: a) Form the equation
Where 
is a parameter.
b) Form the equation using partial differentiation is
(We always try to eliminate
)
c) Solve the all above equation with the given relation 
These give the value of 
These value obtained when substituted in the given function will give the stationary value of the function.
Example1 Divide 24 into three parts such that the continued product of the first, square of second and cube of third may be maximum.
Let first number be x, second be y and third be z.
According to the question
Let the given function be f
And the relation 
By Lagrange’s Method
….(i)
Partially differentiating (i) with respect to x,y and z and equate them to zero
….(ii)
….(iii)
….(iv)
From (ii),(iii) and (iv) we get
On solving
Putting it in given relation we get
Or 
Or 
Thus the first number is 4 second is 8 and third is 12
Example2 The temperature T at any point 
in space is 
.Find the highest temperature on the surface of the unit sphere.
Given function is 
On the surface of unit sphere given 
[
is an equation of unit sphere in 3 dimensional space]
By Lagrange’s Method
….(i)
Partially differentiating (i) with respect to x, y and z and equate them to zero
or
…(ii)
or
…(iii)
…(iv)
Dividing (ii) and (ii) by (iv) we get
Using given relation 
Or 
Or 
So that 
Or 
Thus points are 
The maximum temperature is 
Example3 If
,Find the value of x and y for which 
is maximum.
Given function is 
And relation is 
By Lagrange’s Method
[
] ..(i)
Partially differentiating (i) with respect to x, y and z and equate them tozero
Or 
…(ii)
Or 
…(iii)
Or 
…(iv)
On solving (ii),(iii) and (iv) we get
Using the given relation we get 
So that 
Thus the point for the maximum value of the given function is 
Example-4 Find the points on the surface 
nearest to the origin.
Let 
be any point on the surface, then its distance from the origin 
is
Thus the given equation will be
And relation is 
By Lagrange’s Method
….(i)
Partially differentiatig (i) with respect to x, y and z and equate them to zero
Or 
…(ii)
Or 
…(iii)
Or 
Or 
On solving equation (ii) by (iii) we get
And 
On subtracting we get
Putting in above 
Or 
Thus 
Using the given relation we get
= 0.0 +1=1
Or 
Thus point on the surface nearest to the origin is 
