Unit - 1
Functions of several variables
Q1) What do you understand by the functions of two variables?
A1)
R denotes the set of real numbers. Let’s D is a collection of pairs of real numbers (x , y), which means D is a subset of R × R.
Then a real valued function of two variables of f is a rule that assign to each point (x, y) in D a unique real number denoted by f(x, y).
The set D is called the domain of f.
The set [ f(x, y): (x, y) belongs to D ], which is the set of values the function f takes, is called range of f.
Here we generally use the letter z to denote the value that a function of two variables takes,
Then we will have,
z = f(x, y)
Here we will call that z is the dependent variable and x and y are independent variables.
Q2) Evaluate 
A2)
-6.
Q3) Evaluate 
A3)
Q4) Evaluate 
A4)
1. 
2. 
Here f1 = f2
3. Now put y = mx, we get
Here f1 = f2 = f3
Now put y = mx²
4. 
Therefore,
F1 = f2 = f3 =f4
We can say that the limit exists with 0.
Q5) Evaluate the following-
A5)
First we will calculate f1 –
Here we see that f1 = 0
Now find f2,
Here , f1 = f2
Therefore the limit exists with value 0.
Q6) Calculate 
and 
for the following function
f(x , y) = 3x³-5y²+2xy-8x+4y-20
A6)
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.
Q7) Calculate 
and 
for the following function
f( x, y) = sin(y²x + 5x – 8)
A7)
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)
Q8) If,
Then find.
A8)
We have
Q9) If z = 
Then prove that-
A9)
Here we have-
Then
Q10) What is chain rule for partial derivatives?
A10)
Suppose u = f (x, y) and x and y are themselves functions of a single independent variable
t. Then the dependent variable f may be considered as truly a function of the one independent variable t via the intermediate variables x, y. Now the derivative of f with respect to ‘t ’ is known as the total derivative of f and is given by
Note- if u = f (x, y, z) and x = x(t ), y = y(t ), z = z(t ) then the total derivative of f is
And we can find the this result for more than three variables in same way.
Q11) if w = x² + y – z + sint and x + y = t, find
(a) 
y,z
(b) 
t, z
A11)
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
Q12) If φ(cx – az, cy – bz) = 0 then show that ap + bq = c
Where p = 
q = 
A12)
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
Q13) If z is the function of x and y , and x = 
, y = 
, then prove that,
A13)
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
Q14) State and prove the sufficient condition for differentiability.
A14)
Statement- Suppose 
there exists a neighborhood N(say) of (a, b) at every point of which 
exists. We take (a + h, b + k), a point of this neighborhood so that (a + h, b),(a, b + k) also belongs to N.
Proof:
Since ∂f /∂y is continuous at (a, b), there exists a neighborhood N(say) of (a, b) at every point of which 
exists. We take (a + h, b + k), a point of this neighborhood so that (a + h, b),(a, b + k) also belongs to N
We write f(a + h, b + k) − f(a, b) = {f(a + h, b + k) − f(a + h, b)} + {f(a + h, b) − f(a, b)}. Consider a function of one variable φ(y) = f(a + h, y).
Since 
exists in N, φ(y) is differentiable with respect to y in the closed interval [b, b+k] and as such we can apply Lagrange’s Mean Value Theorem, for function of one variable y in this interval and thus obtain
φ(b + k) − φ(b) = kφ’ (b + θk), 0 < θ < 1
= k
(a + h, b + θk)
f(a + h, b + k) − f(a + h, b) = k
(a + h, b + θk), 0 < θ < 1.
Now, if we write 
(a + h, b + θk) − 
(a, b) = 
(a function of h, k)
Then from the fact that 
is continuous at (a,b). We may obtain
→ 0 as (h, k) → (0, 0).
Again because 
exists at (a, b) implies
f(a + h, b) − f(a, b) = h
(a, b) + 
,
Where 
→ 0 as h → 0. Combining all these we get
Where 
, 
are functions of (h, k) and they tend to zero as (h, k) → (0, 0).
Hence, f(x, y) is differentiable at (a, b).
Q15) If 
where 
then find the value of 
?
A15)
Given 
Where 
By chain rule
Now substituting the value of x ,y,z we get
-6
8
Q16) If
then calculate 
A16) Given 
By Chain Rule
Putting the value of u = 
Again partially differentiating z with respect to y
By Chain Rule
by substituting value 
Q17) If 
where the relation is 
.
Find the value of 
A17)
Let the given relation is denoted by 
We know that 
Differentiating u with respect to x and using chain rule
Q18) Find the directional derivative of 1/r in the direction 
where 
A18)
Here 
Now,
And 
We know that-
So that-
Now,
Directional derivative = 
Q19) Find the directional derivatives of 
at the point P(1, 1, 1) in the direction of the line 
A19)
Here
Direction ratio of the line 
are 2, -2, 1
Now directions cosines of the line are-
Which are 
Directional derivative in the direction of the line-
Q20) Prove that 
A20)
First we will take left hand side
L.H.S = 
= 
= 
= 
Now taking R.H.S,
R.H.S. = 
= 
= 
Here- L.H.S. = R.H.S.
Q21) If f and g are two scalar point functions, then prove that
A21)
L.H.S 
Hence proved
Q22) If 
, then show that
1. 
2. 
A22)
Suppose 
and 
Now taking L.H.S,
Which is 
Hence proved.
2. 
So that
