Unit-1
Partial differentiation and its Applications
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 respect 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
Euler’s theorem
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.
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,
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.
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.
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 = 
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
……
= 
……..
Let u = f(x , y) then the total differential of u which is denoted by du and defined as-
If 
and 
are increments in x and y respectively then the total increment 
in u is given by,
Or
But
Now from first equation-
By using equation (3) in (2) we get the approximation formula-
.......... (iv)
We obtain the approximate value of the function by (4)
Hence
And
Example: Calculate the percentage increase in the pressure p corresponding to a reduction
Of 1/2 % in the volume V, if the p and V are related by 
= C, where C is a constant.
Sol.
Here we have-
= C
Taking log on both sides,
Hence increase in the pressure p = 0.7 percent
Example: Find the approximate value of –
Sol.
Let
Then-
And
Now
Let
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 (
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 
.
- Decompare 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
A differential equation involving partial derivatives with respect to more than one independent variable is called a partial differential equation.
The independent variables will be denoted by x and y and the dependent variable by z. The partial differential coefficients are denoted as-
ORDER of a partial differential equation is the same as that of the order of the highest differential coefficient in it.
Classification of partial differential equation-
Suppose the equation is-
Here A, B, C are the constants of x and y, then the equation-
1. Elliptical- if 
2. Parabolic- if 
3. Hyperbolic- if if
Formation of partial differential equation-
Method of elimination of arbitrary constants-
Example: Form a partial differential equation from-
Sol.
Here we have-
It contains two arbitrary constants a and c
Differentiate the equation with respect to p, we get-
Or
Now differentiate the equation with respect to q, we get-
Now eliminate ‘c’,
We get
Now put z-c in (1), we get-
Or
The second method we use is method of elimination of arbitrary function.
Solution of partial differential equation by direct partial Integration-
Example: Solve-
Sol.
Here we have-
Integrate w.r.t. x, we get-
Integrate w.r.t. x, we get-
Integrate w.r.t. y, we get-
Example: Solve the differential equation-
Given the boundary condition that-
At x = 0, 
Sol.
Here we have-
On integrating partially with respect to x, we get-
Here f(y) is an arbitrary constant.
Now form the boundary condition-
When x = 0,
Hence-
On integrating partially w.r.t.x, we get-
Example: Solve the differential equation-
Given that 
when y = 0, and u = 
when x = 0.
Sol.
We have-
Integrating partially w.r.t. y, we get-
Now from the boundary conditions, 
Then-
From which,
It means,
On integrating partially w.r.t. x givens-
From the boundary conditions, u = 
when x = 0
From which-
Therefore the solution of the given equation is-
Solution of partial differential equation by direct partial Integration-
Example: Solve-
Sol.
Here we have-
Integrate w.r.t. x, we get-
Integrate w.r.t. x, we get-
Integrate w.r.t. y, we get-
Example: Solve the differential equation-
Given the boundary condition that-
At x = 0, 
Sol.
Here we have-
On integrating partially with respect to x, we get-
Here f(y) is an arbitrary constant.
Now form the boundary condition-
When x = 0,
Hence-
On integrating partially w.r.t.x, we get-
Example: Solve the differential equation-
Given that 
when y = 0, and u = 
when x = 0.
Sol.
We have-
Integrating partially w.r.t. y, we get-
Now from the boundary conditions, 
Then-
From which,
It means,
On integrating partially w.r.t. x givens-
From the boundary conditions, u = 
when x = 0
From which-
Therefore the solution of the given equation is-
Lagrange’s linear equation in an equation of the type-
Here P, Q and R are the functions of x, y, z and p = 
and q = 
Working steps to solve-
Step-1: Write down the auxiliary equation-
Step-2: Solve the auxiliary equations-
Suppose the two solutions are- u = 
and v = 
Step-3: Then f(u, v) = 0 or u = ∅(v) is the required solution of
Method of multipliers-
Let the auxiliary equation be
L, m, n may be the constants of x, y, z then we have-
L, m, n are selected in a such a way that-
Thus
On solving this differential equation, if the solution is- u = 
Similarly, choose another set of multipliers 
and if the second solution is v = 
So that the required solution is f(u, v) = 0.
Example: Solve-
Sol.
We have-
Then the auxiliary equations are-
Consider first two equations only-
On integrating
…….. (2)
Now consider last two equations-
On integrating we get-
…………… (3)
From equation (2) and (3)-
Example: Find the general solution of-
Sol. The auxiliary simultaneous equations are-
……….. (1)
Using multipliers x, y, z we get-
Each term of (1) is equals to-
Xdx + ydy + zdz=0
On integrating-
………… (2)
Again equation (1) can be written as-
Or
………….. (3)
From (2) and (3), the general solution is-
Non-linear partial differential equations-
Type-1: Equation of the type f(p, q) = 0
Method-
Let the required solution is-
Z = ax + by + c …….. (1)
So that-
On putting these values in f(p, q) = 0
We get-
f(a, b) = 0
So from this, find the value of b in terms of a and put the value of b in (1). It will be the required solution.
Type-2: Equation of the type-
Z = px + qy + f(p, q)
Its solution will be-
Z = ax + by + f(a, b)
Type-3: Equation of the type f(z, p, q) = 0
Type-4: Equation of the type-
Method-
Let-
Example: Solve-
Sol.
This equation can be transformed as-
………. (1)
Let
Equation (1) can be written as-
………… (2)
Let the required solution be-
From (2) we have-
Example: Solve-
Sol.
Let u = x + by
So that-
Put these values of p and q in the given equation, we get-
Example: Solve-
Sol.
Let-
That means-
Put these values of p and q in
This is the general method for finding the complete integral of a non-linear partial differential equation.
Let us consider the equation-
f(x, y, z , p, q) = 0 ………. (1)
Since z depends on x and y, we have-
The main thing in Charpit’s method is to find the another relationship between the variables x, y, z and p. q.
And let the relation be-
Here on solving equation (1) and (2), we get the values of p and q.
When we substitute these values of p and q in (2), it becomes integrable.
To determine 
, equation (1) and (3) are differentiated with respect to x and y.
We get, when we differentiate with respect to x,
We get, when we differentiate with respect to y,
Eliminating 
between the equations we get from differentiating for x, we get
Or
…………(4)
Eliminating 
between the equations we get from differentiating for y, we get
…………(5)
Adding (4) and (5) and keeping in view the relation on, the terms of the last brackets of (4) and (5) cancel. On rearranging, we get-
Or
This equation is the Lagrange’s linear equation of the first order with x, y, z, p, q as independent variables and equation (4) as dependent variable.
Its subsidiary equations are-
An integral of these equations involving p or q or both can be taken as the relation (3) which along with (1) will give the values of p and q to make (2) integrable.
Example: Solve-
Sol.
Let
Charpit’s subsidiary equations are-
So that- dq = 0 or q = a
On putting q = a in (1) we get-
Such that-
Integrating
Or
Which is the required solution.
Example: Solve-
Sol.
Let
Charpit’s subsidiary equations are-
From the first and fourth ratios,
Substituting p = a – x in the given equation, we get-
So that-
Multiply both sides by 
,
Integrating-
Or
Which is the required solution.
