Course: Accelerated Engineering Calculus II Instructor: Michael Medvinsky 4.3 Partial Derivatives (11.3)
Def: A partial derivative of a function of several variables is a derivative with respect of one of those variables why the other variables consider constant. For a function f ( x, y ) , the partial derivatives are: f ( x + h, y ) − f ( x, y )
∂
Dx f =
f ( x, y ) = fx ( x, y ) = lim
h→0
∂x
h
f ( x, y + h ) − f ( x, y )
∂
Dy f =
f ( x, y ) = fy ( x, y ) = lim
h→0
∂y
h
For a function f ( x, y, z ) , the partial derivatives are: f ( x + h, y, z ) − f ( x, y, z )
∂
f ( x, y, z ) = fx ( x, y, z ) = lim
h→0
∂x
h
f ( x, y + h, z ) − f ( x, y, z )
∂
Dy f =
f ( x, y, z ) = fy ( x, y, z ) = lim
h→0
∂y
h
Dx f =
f ( x, y, z + h ) − f ( x, y, z )
∂
f ( x, y, z ) = fz ( x, y, z ) = lim
h→0
∂z
h
Similarly, for a function f ( x1 , x2 ,...xn ) the partial derivatives D1 f = f1 , D2 f = f2 ,..., Dn f = fn Dz f =
Ex 1.
f x′ = -
Find partial derivatives of f (x, y) =
sin(xy)yx 2 +sin(xy)y3 +2cos(xy)x
(x 2 +y2 )2
f y′ = -
cos(xy)
x2 + y2
sin(xy)xy2 +sin(xy)x 3 +2cos(xy)y
(x 2 +y2 )2
Ex 2.
Verify whether the partial derivatives of
⎧ xy
⎪
f ( x, y ) = ⎨ x 2 + y 2
⎪0
⎩
( x, y ) ≠ (0,0)
( x, y ) = (0,0)
continuous or not.
The function f ( x, y ) is continuous at every ( x, y ) ≠ ( 0,0 ) as a fractional function, it is also continuous at ( x, y ) = ( 0,0 ) since lim
( x,y )→(0,0)
r 2 cosθ sin θ
= lim
= lim r cosθ sin θ = 0 = f ( 0,0 ) ( x,y )→(0,0)
r
x 2 + y 2 ( x,y )→(0,0)
xy
Next we find partial derivatives at ( x, y ) = ( 0,0 ) f (h,0) − f (0,0)
h ⋅0 / h 2
f (0,h) − f (0,0)
0 ⋅ h / h2
= lim
= 0 and fy (0,0) = lim
= lim
=0
h→0
h→0
h→0
h→0
h
h
h
h
For x, y ≠ 0,0 : fx (0,0) = lim
( ) ( )
48 Course: Accelerated Engineering Calculus II Instructor: Michael Medvinsky y x 2 + y 2 − xy
f x (x, y) =
x2 y + y3 − x2 y
2x
2 x2 + y2
x2 + y2
x x 2 + y 2 − xy
f y (x, y) =
x2 + y2
x2 + y2
=
=
(x
y3
2
2
)
+ y2
)
+y
3/2
x 3 + xy 2 − y 2 x
2y
2 x2 + y2
x2 + y2
x2 + y2
x2 + y2
=
=
(x
x3
2
3/2
Finally, fx , fy aren’t continuous at ( x, y ) = ( 0,0 ) since x3
1
lim fx (x, y) = lim fx (x, x) = lim
= 3/2 ≠ 0 = fx ( 0,0 )
3/2
( x,y)→(0,0)
( x,x )→(0,0)
x→0
2
2x 2
( )
lim
( x,y)→(0,0)
fy (x, y) =
lim
( x,x )→(0,0)
fy (x, x) = lim
x→0
x3
( 2x )
2 3/2
1
= 3/2 ≠ 0 = fy ( 0,0 )
2
4.3.1 Higher derivatives
Partial derivatives of a function of several variables are functions of several variables, and therefore, may have their partial derivatives. We denote it as: fxx =
∂f
∂x 2
fxy =
∂ ⎛∂f ⎞
∂f
⎜⎝ ⎟⎠ =
∂y ∂x
∂y ∂x
fyx =
∂ ⎛∂f ⎞
∂f
=
⎜
⎟
∂x ⎝ ∂y ⎠ ∂x ∂y
fyy =
∂f
∂y 2
Thm: If fxy , fyx continuous around ( a,b ) then fxy , ( a,b ) = fyx ( a,b ) Ex 3.
y
x +y
+ y+
. Find f x , f y , f xx , f yy , f xy , f yx
2
2
x
y2
2y
f x (x, y) = −2 3 + x
f y (x, y) = 2 + 1+ y
x
x
2
y
2
f xx (x, y) = 6 4 + 1
f yy (x, y) = 2 + 1
x
x
y
y
f xy (x, y) = −4 3
f yx (x, y) = −4 3
x
x
Let f (x, y) =
2
2
2
4.3.2 Partial differential equations
Def: Partial differential equations are differential equations that involve partial derivatives. Ex 4. Find k such that u = sin ax sinby satisfies Helmholtz equation u xx + u yy + k 2u = 0 .
u x = a cos ax sinby
u xx = −a 2 sin ax sinby = −a 2u
u y = bsin ax cosby
u yy = −b 2 sin ax sinby = −b 2u
u xx + u yy = − ( a 2 + b 2 ) u ⇒ k 2 = a 2 + b 2
49