FUNDAMENTALS OF BUSINESS MATHEMATICS AND STATISTICS I 3.41
Now differentiating u w.r.t.x (treating y as a constant) we find
fy u (x 4xy 3y ) (x ) (4xy) (y ) 2x 4y 0 2x 4y.^2222
x x x x x
∂ =∂ + + = ∂ +∂ +∂ = + + = +
∂ ∂ ∂ ∂ ∂
Again (treating x as a constant)
fy u (x 4xy 3y )^22 (x ) (4xy) (y )^22
y y y y y
∂ = ∂ + + = ∂+ + ∂ + ∂
∂ ∂ ∂ ∂ ∂
= 0 + 4x + 2y = 4x + 2y.
Example 89: If u = x^4 y^3 z^2 + 4x + 3y + 2z find ∂∂u u ux y z, , .∂∂ ∂∂
Solution:
u 4x y z 4 3 3 2
x
∂ = +
∂ ,
u 3y x z 3 2 4 2
y
∂ = +
∂ ,
u 2zx y 2. 4 3
z
∂ = +
∂
Example 90: If u = log (x^2 + y^2 ); find
2 2
2 2
u u,.
x y
∂ ∂
∂ ∂
2 2
u 1 .2x
x x y
∂ =
∂ + ,
2 2 2 2 2
2 2 2 2 2 2 2
u (x y ).2 2x(2x) 2(y x )
x (x y ) (x y )
∂ = + − = −
∂ + +
( ) ( )
( )
( )
( )
2 2 2 2 2
2 2 2 2 22 2 22
u 2y , u x y 2 2y 2y 2 x y
y x y y x y x y
∂ = ∂ = + − = −
∂ + ∂ + +
Example 91:
For f(x, y) = 3x^2 – 2x + 5, find fx and fy
Solution:
fx = 6x – 2 and fy = 0.
Example 92 :
If f(x, y) = 3 3
x y
y x− , find fx and fy
Solution:
x 3 4
f 1 3y
=y x+ y 4 3
f 3x 1
= −y x−
Example 93:
Find the first partial derivatives of f(x, y, z) = xy^2 z^3
Solution:
fx = 2 – u fy = z + 2uy fx = 3xy^2 z^2