Module-2
Differential calculus-2
Taylor’s series-
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-
+ …….+ 
+ ……..
Is called Taylor’s series.
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-
Example-2: Expand 
in powers of (x – 1).
Sol.
Here we have-
Now-
Put these values in Taylor’s theorem-
We get-
Example-3: By using Maclaurin’s series expand tan x.
Sol.
Let-
Put these values in Maclaurin’s series we get-
Example-4: Expand 
by using Maclaurin’s series.
Sol.
Let
Put these values in Maclaurin’s series-
Or
Let we have two functions f(x) and g(x) and-
Then-
Is an expression of the form 
In that case we can say that f(x)/g(x) is an indeterminate for of the type 
at x = a.
Now, Let we have two functions f(x) and g(x) and-
Then-
Is an expression of the form 
, in that case we can say that f(x)/g(x) is an indeterminate for of the type 
at x = a.
Some other indeterminate forms are 
L’Hospital’s rule for 
form-
Working steps-
1. Check that the limits f(x)/g(x) is an indeterminate form of type 
.
(Note- we can not apply L’Hospital rule if it is not in indeterminate form)
2. Differentiate f and g separately.
3. Find the limits of the derivatives .if the limit is finite ,
then it is equal to the limit of f(x)/g(x).
Example-1: Evaluate 
Sol. Here we notice that it is an indeterminate form of 
.
So that , we can apply L’Hospital rule-
Example-2: Evaluate 
.
Sol. Let f(x) = 
and g(x) = 
.
Here we see that this is the indeterminate form of 0/0 at x = 0.
Now by using L’Hospital rule, we get-
= 
= 
= 
= 1
Note- Suppose we get an indeterminate form even after finding first derivative, then in that case , we use the other form of L’Hospital’s rule.
If we have f(x) and g(x) are two functions such that 
.
If 
exist or (∞ , -∞), then
Example-3: Evaluate 
Sol. Let f(x) = 
, then
And
= 0
= 0
But if we use L’Hospital rule again, then we get-
Example-4: Evaluate 
Sol. We can see that this is an indeterminate form of type 0/0.
Apply L’Hospital’s rule, we get
But this is again an indeterminate form, so that we will again apply L’Hospital’s rule-
We get
= 
L’Hospital’s rule for 
form-
Let f and g are two differentiable functions on an open interval containing x = a, except possibly at x = a and that
If 
has a finite limit, or if it is 
, then
Theorem- If we have f(x) and g(x) are two functions such that 
.
If 
exist or (∞ , -∞), then
Example-5: Find 
, n>0.
Sol. Let f(x) = log x and g(x) = 
These two functions satisfied the theorem that we have discussed above-
So that,
Example-6: Evaluate 
Sol. Apply L’Hospital rule as we can see that this is the form of 
= 
Note- In some cases like above example, we can not apply L’Hospital’s rule.
Other types of indeterminate forms-
Example-7: Evaluate 
Sol. Here we find that-
So that this limit is the form of 0
.
Now,
Change 
to obtain the limit-
Now this is the form of 0/0,
Apply L’Hospital’s rule-
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
Hence proved.
Higher order partial derivatives-
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 series-
+ …….+ 
+ ……..
2. Maclaurin’s Theorem-
+ …….
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 + sin t 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.
Derivatives of composite and implicit functions (Chain rule) -
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 cos xy
Fy = 2y + x cos xy
So ,
Example-2: Let 
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,
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 
Example-1: Find out the maxima and minima of the function
Sol.
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 (
Example-3: There is rectangular box which is open at the top, is to have a volume of 32 c.c.
Find the dimensions of the box requiring least (minimum) material to construct it.
Sol. Here it is given that-
Volume (V) = 32 c.c.
Suppose ‘l’ , ‘b’, ‘h’ are the length, breadth, height of the rectangular box respectively and its surface area is ‘S’.
As we know that-
Volume (V) = l b h = 32
b (breadth) = 32/lh
And the surface area of the rectangular box which is open at the top is-
S = 2 (l + b) h + l b …………… (1)
On putting the value of ‘b’ in (1), we get-
S = 2 
S = 
……………… (2)
Differentiate partially equation (2) with respect to l and h respectively, we get-
…………… (3) and
…………….. (4)
For Max. And Min. S, we get-
And 
The values of ‘l’, ‘h’ and ‘b’ will be-
L = 4, h = 2 and b = 4
Now-
And
So that-
, then S is minimum for l = 4, b = 4, and h = 2
Key takeaways-
- A point is a saddle point of a function of two variables if
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@f-= 0, @f--= 0, and @-f- @-f-- -@-f- < 0
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At the point.
2. The value 
is said to be a stationary value of 
if
Suppose 
is a function of three independent variables, where x,y,z are related to a constraint
(x, y, z) = 0 …………………… (1)
So that for stationary values-
, 
, 
………………….. (2)
By total differentiation of (1), we get-
…………………… (3)
Now multiply by 
and adding to (2), we get-
That equation holds good if-
We can find the values of x, y, z and 
by solving these equations.
Example-1: Determine the maxima and minima of 
when 
1
Sol. Suppose 
……….. (1)
And 
=0 …………… (2)
Differentiate partially equation (1) and (2) w.r.t. x and y respectively, we get-
And 
Now using Lagrange’s equations, we get-
Which gives-
or
……… (3)
Which gives-
…… (4)
Which gives-
……. (5)
Multiply these equations by x, y, z respectively and adding, we get-
(
Hence we get-
f + 
0 then 
Put
in (3) , (4) and (5), we get-
x(1 – fa) = 0 , y(1 – fb) = 0 and z(1 – fc) = 0
We get-
These gives the maximum and minimum values of f.
Example: 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 
.
Example: 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
Maxima and minima of function of two variables and three variables using method of Lagrange’s multiplier-
Generally we 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 fails 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 
We use the concept of maxima and minima to find the maximum and minimum value of the function.
Example: Find the volume of the greatest rectangular parallelepiped that can be inscribed in the ellipsoid-
Sol.
Suppose the edges of the parallelepiped are 2x, 2y, 2z and these are parallel to the axes
Then we know that the volume will be-
V = 8xyz
Here we need to find out the maximum value of V with the given conditions-
Write-
Then
And
And
From equation (1) and (2), equate the value of 
, we get-
Similarly from (2) and (3), we get-
We can say that
Put these values in equation (A),
We get-
Which means-
Which gives-
When x = 0, the parallelepiped is a rectangular sheet and its volume V = 0.
As x increases, V also increases.
So that V must be maximum (stage given by eq-4)
Hence the greatest (Maximum) volume is-
Example: Show that the rectangular solid of maximum volume that can be inscribed in a sphere is a cube.
Sol.
Suppose 2x, 2y, 2z are the length, breadth and height of the rectangular solid.
Its volume will be-
V = 8xyz ....... (1)
And let R is the radius of the sphere so that-
Then
Now
Or
So that for a maximum value x = y = z
Now we can say that the rectangular solid of maximum volume that can be inscribed in a sphere is a cube.
Example-2: Find the minimum distance from the point (1, 2, 0) to the cone 
Sol. Let there are three points (x, y, z) on the cone then the distance from the point (1, 2, 0) by using distance formula will be-
Suppose 
………….. (1)
Subject to 
………………. (2)
From equation (1) and (2), for minimum, we get-
………….. (3)
……………. (4)
Multiply (4) by 
and adding in (3), we get-
, 
, 
We get-
We get the values of x and y-
x = 1/2 , y = 1
Put the values of x and y in equation (2), we get-
So that the min. Distance form the point (1, 2, 0) is-
Therefore,
Ans.
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
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
= 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
Jacobian of Implicit functions-
If the variables u,v, x, and y are connected by the following equations-
………………… (1)
………………….. (2)
u,v are the implicit functions of x and y.
Now differentiate equation (1) and (2) with respect to x and y-
………… (3)
……………… (4)
………………. (5)
………………. (6)
Now,
= 
Using equation (3), (4), (5), (6), we get-
= 
So that-
We can find this for three variables as well.
Example-1: If u = 2axy, v = 
then prove that-
Sol. Here we have,
u = 2axy, v = 
Then
Here - 
So that
Now,
Hence-
Hence proved.
Example-2: If 
, then prove that-
Sol. Suppose 
And 
Now
And
So that,
Hence proved.
Example-3: If x + y + z = u, y + z = uv , z = uvw, then prove that-
Sol. Suppose
And
Hence,
Hence proved.
Jacobians for functional dependence functions-
Note- two functions u and v said to be functionally dependent if their jacobian is equals to zero. That means J (u , v) = 0
Suppose u and v are functionally dependent functions, then
f( u, v) = 0
Differentiate this equation with respect to x and y-
= 0
= 0
There will be a non-trivial solution for 
, 
to this system exists.
So that-
On transposing, we get-
Example-1: Show that 
and 
are functionally dependent.
Sol. Here we have-
and
Now we will find out the Jacobian to check the functional dependence.
= 
Here Jacobian is zero, so we can conclude that these functions are functionally dependent.
Example-2: Prove that u, v, w are functionally dependent, where-
Sol. Here we have
Now we will find out the Jacobian of the given functions-
= 
= 
= 0
Therefore, u, v, w are functionally dependent.
Key takeaways-
- 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,
2. If u and v are the functions of x and y , then
3. Suppose u and v are the functions of r and s, where r,s are the fuctions of x , y, then,
4. If u,v,w are the functions of three independent variables x,y,z are not independent , then,
References:
- C.Ray Wylie, Louis C.Barrett : "Advanced Engineering Mathematics", 6th Edition, 2. McGraw-Hill Book Co., New York, 1995.
- James Stewart : “Calculus -Early Transcendentals”, Cengage Learning India Private Ltd., 2017.
- B.V.Ramana: "Higher Engineering Mathematics" 11th Edition, Tata McGraw-Hill, 2010.
- Srimanta Pal & Subobh C Bhunia: “Engineering Mathematics”, Oxford University Press, 3rd Reprint, 2016.
- Gupta C.B., Singh S.R. And Mukesh Kumar: “Engineering Mathematics for Semester I & II”, Mc-Graw Hill Education (India) Pvt.Ltd., 2015.
