Unit-1
Partial differentiation and its Applications
Question Bank
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: 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
Question-3: 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.
Question-4: 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.
Question-5: 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,
Question-6: 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
Question-7: 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-8: 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-9: 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 = 
Question-10: 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
……
= 
……..
Question-11: 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
Question-12: 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 (
Question-13: 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
Question-14: 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-
Question-15: 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-
Question-16: 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)-
Question-17: Solve-
Sol.
Let u = x + by
So that-
Put these values of p and q in the given equation, we get-
Question-18: 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.
