Unit-4
Multivariable Differential Calculus
Question-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.
Question-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)
Question-3: Find
Sol. First we will differentiate partially with repsect to r,
Now differentiate partially with respect to θ, we get
Question-4: 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.
Question-5: 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
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
Question-6: 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.
Question-7: 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.
Question-8: Show that 
Given 
Therefore f(x,y,z) is an homogenous equation of degree 2 in x, y and z
Question-9: If
Let 
Thus u is an homogenous function of degree 2 in x and y
Therefore by Euler’s theorem
substituting the value of u
Hence proved
Question-10: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
Unit-4
Multivariable Differential Calculus
Question-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.
Question-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)
Question-3: Find
Sol. First we will differentiate partially with repsect to r,
Now differentiate partially with respect to θ, we get
Question-4: 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.
Question-5: 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
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
Question-6: 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.
Question-7: 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.
Question-8: Show that 
Given 
Therefore f(x,y,z) is an homogenous equation of degree 2 in x, y and z
Question-9: If
Let 
Thus u is an homogenous function of degree 2 in x and y
Therefore by Euler’s theorem
substituting the value of u
Hence proved
Question-10: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
