Paper 4: Fundamentals of Business Mathematics & Statistic

3.40 I FUNDAMENTALS OF BUSINESS MATHEMATICS AND STATISTICS

Calculus

Example 85 : u = (x + y)^2 = x^2 + 2xy + y^2 ,

u 2x 2y, 2x 2y.u

x y

∂ = + ∂ = +

∂ ∂

Function of three variables : A function may be of three variables also, i.e., u = f(x, y, z). Now the partial

derivative of u w.r.t. x is the derivative of u w.r.t.x (treating y and z as constant).

Example 86 : u = x^2 + y^2 + z^2 ;

u 2x, u 2y, 2z.u

x y z

∂ = ∂ = ∂ =

∂ ∂ ∂

Partial derivative of higher order

Partial derivative of higher order is obtained by usual method of derivative.

For the function u = f(x, y), we have following four partial derivative of second order :

(i)

2

2 xx

u u f

x x x

∂ = ∂ ∂ (^) =
∂ ∂^ ∂^ (ii)
2
2 yy
u u f
y y y
∂ = ∂ ∂ (^) =
∂ ∂^ ∂^
(iii)
2
yx
u u f
x y x y
∂ = ∂ ∂ (^) =
∂ ∂ ∂^ ∂^ (iv)
2
xy
u u f.
y x y x
∂ = ∂ ∂ (^) =
∂ ∂ ∂^ ∂^
Example 87 : u = 2x^2 – 4xy + 3y^2 find
2 2 2
2
u u u, ,
x y x x y

∂ ∂ ∂

∂ ∂ ∂ ∂ ∂

( )

2

2

u 4x 4y, u u 4x 4y 4.

x x x x x

∂ = − ∂ = ∂ ∂ (^) = ∂ − =
∂ ∂ ∂^ ∂^ ∂
Again ( )
(^2) u u
∂ ∂∂y x y x y=∂∂^ ∂∂^ =∂∂ 4x 4y 4.− = −
( )
u^2 u u
∂∂y= −4x 6y,+ ∂ ∂∂x y x y x=∂∂ ∂∂ (^) =∂∂ −4x 6y 4+ = −
Note : (i)
(^2) u
y x

∂

∂ ∂ means partial derivative of

u

x

∂

∂ w.r.t.y.

(ii)

(^2) u
x y

∂

∂ ∂ means partial derivative of

u

y

∂

∂ w.r.t.x.

(iii) We get

(^2) u (^2) u
y x x y

∂ = ∂

∂ ∂ ∂ ∂ (from the above result) i.e., ux y

(^) = uyx.
Example 88: Find partial derivative of first order of the function x^2 + 4xy + y^2
Solution:
Let u = f(x, y) = x^2 + 4xy + y^2.