Partial Derivatives
Mathematics 54 - Elementary Analysis 2
Institute of Mathematics University of the Philippines-Diliman
Partial Derivatives
Mathematics 54 - Elementary Analysis 2
Introduction
Letz=f(x,y) be defined at (a,b). If we fixy=b, thenf(x,b) becomes a function ofxalone.
The slope of the above tangent line to this curve at (a,b) is given by d
dx
¡
f(x,b)¢¯¯ ¯x
=a=:fx(a,b).
Similarly, if we fixx=a, thenf(a,y) becomes a function ofyalone.
The slope of the above tangent line to this curve at (a,b) is given by d
dy
¡
f(a,y)¢¯ ¯ ¯
y=b=:fy(a,b).
Partial Derivatives
Definition
Letz=f(x,y).
1 Thepartial derivative offwith respect toxat any point (x,y) is
fx(x,y)= lim
∆x→0
f(x+∆x,y)−f(x,y)
∆x ,
provided that this limit exists.
2 Thepartial derivative offwith respect toyat any point (x,y) is
fy(x,y)= lim
∆y→0
f(x,y+∆y)−f(x,y)
∆y ,
provided that this limit exists.
Example
Findfx(x,y) given thatf(x,y)=2x2y2+2y+4x.
Solution.Consider
f(x+∆x,y)−f(x,y)
∆x =
2(x+∆x)2y2+2(y)+4(x+∆x)−(2x2y2+2y+4x) ∆x
= 2x
2y2+4x(∆x)y2+2(∆x)2y2+2y+4x+4∆x−2x2y2−2y−4x
∆x
= 4x(∆x)y
2+2(∆x)2y2+4∆x
∆x = 4xy2+2y2∆x+4
Thus,
fx(x,y) = lim
∆x→0(4xy 2
+2y2∆x+4)
Partial Derivatives
Remark.
The partial derivatives off at (a,b) are also given by
1 f
x(a,b)=lim x→a
f(x,b)−f(a,b) x−a
2 fy(a,b)=lim
y→b
f(a,y)−f(a,b) y−b
Example
Findfx(1, 1) andfy(1, 1) given thatf(x,y)=
3x y +4xy
2.
Solution.Forfx, let us use the definition. We have
f(1+∆x, 1)−f(1, 1)
∆x =
3(1+∆x)
1 +4(1+∆x)(1) 2−
µ3(1)
1 +4(1)(1) 2
¶
∆x
= 3+3∆x+4∆x−3 ∆x
= 7∆x ∆x
= 7.
Thus,
fx(1, 1)= lim
Example.(continued...)
Findfx(1, 1) andfy(1, 1) given thatf(x,y)=
3x y +4xy
2.
Forfy, let us use the previous remark. Consider
f(1,y)−f(1, 1) y−1 =
3 y+4y
2−7
y−1
=
3+4y3−7y y y−1
= (2y+3)(2y−1)(y−1)
y(y−1)
= (2y+3)(2y−1)
y .
Hence,
fy(1, 1)=lim y→1
(2y+3)(2y−1) y =5.
Remark.
The partial derivative with respect to a variable can also be obtained by applying theorems for ordinary differentiation while treating the other variables constant.
Example
Determine the partial derivative off(x,y)=2x2y2+2y+4xwith respect to x.
fx(x,y) = (2y2)
d dx(x
2) +2y d
dx(1)+4 d dx(x)
Partial Derivatives
Example
Findfx(1, 1) andfy(1, 1) given thatf(x,y)=
3x y +4xy
2.
First,fx(x,y)=
3 y
d dx(x)+4y
2 d dx(x)=
3 y+4y
2.
Thus,fx(1, 1)=
3 1+4(1)
2 =7.
Similarly,fy(x,y)=3x
d dy
µ1
y
¶ +4x d
dy(y 2)
=−y3x2 +8xy.
Hence,fy(1, 1)=−
3(1)
12 +8(1)(1)=5.
Notations.
Letz=f(x,y). The following denotes the partial derivatives off with respect tox:
fx(x,y)
Dx ¡
f(x,y)¢
∂ ∂x
¡
f(x,y)¢
∂z ∂x
Similarly, the partial derivatives off with respect toyare denoted by:
fy(x,y)
Dy ¡
f(x,y)¢
∂ ∂y
¡
f(x,y)¢
Partial Derivatives
Example
Evaluate ∂ ∂ysin
2(2x−3y).
Solution.
∂ ∂ysin
2(2x
−3y) = ∂
∂y
¡
sin(2x−3y)¢2
= 2 sin(2x−3y) cos(2x−3y)(−3)
= −6 sin(2x−3y) cos(2x−3y)
Note.
fxandfyare also called the derivatives off in thex- andy-directions,
respectively. Moreoever, they can also be interpreted as rates of change.
Remark.
For a function ofnvariablesx1,x2, . . . ,xn, there arenpartial derivatives.
The partial derivative with respect to theith variable is
fxi(x1,x2, . . . ,xn)= lim
∆xi→0
f(x1, . . . ,xi−1,xi+∆xi,xi+1, . . . ,xn)−f(x1, . . . ,xn) ∆xi
.
Also,fxi(x1,x2, . . . ,xn) is obtained by performing ordinary differentiation
with respect toxiwhile treating the other variables constant.
Notation.
Letu=f(x1,x2, . . . ,xn). The following are the other notations for the partial
derivative with respect toxi:
Dxi £
f(x1,x2, . . . ,xn) ¤
Partial Derivatives of Functions of
n
Variables
Example
Letf(x,y,z)=xz
y2+sin(2x+3y+4z)+e
x2z. Findf
x,fyandfz.
Solution.
fx(x,y,z) =
z
y2+2 cos(2x+3y+4z)+2xze x2z
fy(x,y,z) = −
2xz
y3 +3 cos(2x+3y+4z)
fz(x,y,z) =
x
y2+4 cos(2x+3y+4z)+x 2ex2z
Letz=f(x,y). The functionsfx(x,y) andfy(x,y) are also functions ofxand
y.
These functions also have partial derivatives. In fact, there are four of them.
∂ ∂x
µ∂z
∂x
¶
= ∂
2z
∂x2=fxx ∂
∂y
µ∂z
∂x
¶
= ∂
2z
∂y∂x=fxy ∂
∂x
µ∂z
∂y
¶
= ∂
2z
∂x∂y=fyx ∂
∂y
µ∂z
∂y
¶
= ∂
2z
∂y2=fyy
The above derivatives are called thesecond order partial derivativesoff.
Higher Order Partial Derivatives
Example
Letw=f(x,y)=ln(2x+3y). Find (a) ∂ 2w
∂x∂y and (b)fxy(x,y).
Solution. (a)∂w
∂y =fy(x,y)= 1 2x+3y(3)=
3 2x+3y
∂2w
∂x∂y=fyx(x,y)= ∂ ∂x
£
3(2x+3y)−1¤
=(−3)(2x+3y)−2(2)
= −(2x 6 +3y)2
(b)∂w
∂x =fx(x,y)= 1 2x+3y(2)=
2 2x+3y
∂2w
∂y∂x=fxy(x,y)= ∂ ∂y
£
2(2x+3y)−1¤
=(−2)(2x+3y)−2(3)= − 6
(2x+3y)2
Remark.The above example shows thatfxy=fyx.
Clairaut’s Theorem
Letf be a function ofxandy. Iffxyandfyxare continuous on some
circular regionRon a plane, then for all (x,y) onR,
fxy=fyx.
Example
No functionf(x,y) exists such thatfx(x,y)=3x+2yandfy(x,y)=4x+5y.
Indeed, if such a function exists, thenfxy=fyx. Butfxy(x,y)=2 and
fyx(x,y)=4, which are both continuous, yet are not equal.
Remark.
Higher Order Partial Derivatives
Example
Letf(x,y,z)=zln(x2ycosz). Find ∂ 3f
∂x∂x∂z andfxzx.
Solution.
∂f(x,y,z)
∂z = ln(x
2ycosz) +z
µ 1
x2ycosz
¶ ¡
x2y(−sinz)¢
= 2 lnx+lny+ln(cosz)−ztanz
∂2f(x,y,z) ∂x∂z =
2 x
∂3f(x,y,z) ∂x∂x∂z = −
2 x2
Forfxzx, recall thatfzx=fxzfor all (x,y,z) in the domain off. Thus,
fxzx=fzxx= −
2 x2.
1 Findfxandfyforf(x,y)=tan(3x−y)+x
y.
2 Findgxandgyforg(x,y)=xy.
3 Findhxandhyforh(x,y)= x
py
x2−y.
4 Findfx,fyandfzforf(x,y,z)=log
x(y)−xz2.
5 Solve for ∂
2f
∂x∂z iff(x,y,z)= x2−3y
z .
