Differentiability, Differentials and Local Linear
Approximation
Mathematics 54 - Elementary Analysis 2
Differentiability
Recall. Ify=f(x), we sayf is differentiable atx=aiff0(a) exists.
However, forz=f(x,y), differentiability at (x,y)=(a,b) is not as simple as the existence offx(a,b) andfy(a,b).
For a functionf(x,y), define
∆f =f(x0+∆x,y0+∆y)−f(x0,y0),
called theincrementoff.
Differentiability
Definition.
A functionf of two variablesxandyis said to bedifferentiableat (x0,y0) if
fx(x0,y0) andfy(x0,y0) both exist and
lim (∆x,∆y)→(0,0)
∆f−fx(x0,y0)∆x−fy(x0,y0)∆y p
(∆x)2+(∆y)2 =0
Remarks.
1 A functionf(x,y) is said to bedifferentiable inR⊆R2if it is differentiable at all points inR.
Differentiable Functions
Theorem.
1 If a function is differentiable at a point, then it is continuous at that point.
2 If all first-order partial derivatives off exist and are continuous at a point, thenf is differentiable at that point.
Differentiability
Examples.
1 The functionf(x,y)=exyis differentiable everywhere since
fx(x,y)=yexyandfy(x,y)=xexyare continuous everywhere.
2 The functionf(x,y)= (
0 ifxy6=0,
1 ifxy=0 is not differentiable at (0, 0) since it is not continuous at (0, 0).
3 The functionf(x,y)=px2+y2is not differentiable at (0, 0) since
fx(x,y)=p x
Differentiability
Example.Letf(x,y)=
0 if (x,y)=(0, 0)
xy
p
x2+y2 if (x,y)6=(0, 0) .
It can be verified, by definition, thatf is continuous at (0, 0).
We check the existence of bothfx(0, 0) andfy(0, 0).
fx(0, 0) = lim x→0
f(x, 0)−f(0, 0)
x−0
= lim x→0
0−0
x
= 0
Similarly,fy(0, 0)=0. Thus, the partial derivatives exist at (0, 0).
Differentiability of
f
(
x
,
y
)
=
0
if (
x
,
y
)
=
(0, 0)
xy
p
x2+y2
if (
x
,
y
)
6=
(0, 0)
Let (x0,y0)=(0, 0). Consider
∆f−fx(x0,y0)∆x−fy(x0,y0)∆y q
(∆x)2+(∆y)2 =
f(0+∆x, 0+∆y)−f(0, 0)−fx(0, 0)∆x−fy(0, 0)∆y q
(∆x)2+(∆y)2
=
(∆x)(∆y) q
(∆x)2+(∆y)2 q
(∆x)2+(∆y)2 = (∆x)(∆y)
(∆x)2+(∆y)2
As (∆x,∆y)→(0, 0) along the line∆x=0, its limit is (∆x,∆y)→(0, 0) is zero. But along the line∆x=∆y, the limit is
lim ∆x→0
(∆x)2 2(∆x)2=
Differentiability of
f
(
x
,
y
)
=
0
if (
x
,
y
)
=
(0, 0)
xy
p
x2+y2
if (
x
,
y
)
6=
(0, 0)
Thus, lim (∆x,∆y)→(0,0)
∆f−fx(x0,y0)∆x−fy(x0,y0)∆y p
Differentiability, Differentials and Local Linear
Approximation
Mathematics 54 - Elementary Analysis 2
Differentials
Differentials
Letz=f(x,y) be differentiable at (x0,y0). We define the(total) differential ofz at (x0,y0) by
dz:=£
fx(x0,y0) ¤
dx+£
fy(x0,y0) ¤
dy
wheredxanddyare the changes in the values ofxandyrespectively.
Remarks.
1 The differential ofz=f(x,y) at (x
0,y0) approximates∆f, which is the change in the value off(x,y) when (x,y) changes from (x0,y0) to (x0+dx,y0+dy).
2 At any (x,y) wherez=f(x,y) is a differentiable ,dzcan also be written as
dz=∂z
∂xdx+
∂z
Differentials
Example
Letz=tan−1(xy). Computedzand comparedzwith the change inzwhen (x,y) changes from (1, 1) to (0.9, 1.01).
Solution.The differential ofzis given by
dz =
·∂
∂xtan
−1(xy)¸dx +
·∂
∂ytan
−1(xy)¸dy
=
µ y
1+x2y2 ¶
dx+
µ x
1+x2y2 ¶
dy
Computing the differentials at (1, 1), we have
dx = 0.9−1= −0.1
dy = 1.01−1=0.01
dz =
µ 1
1+12(1)2 ¶
(−0.1)+
µ 1
1+12(1)2 ¶
(0.01)= −0.045
Differentials in Word Problems
Example
The legs of a right triangle are measured to be 3cm and 4cm, with a maximum error of 0.05cm in each measurement. Use differentials to approximate the maximum possible error in the calculated value of (a) the hypotenuse and (b) the area of the triangle.
Solution.Letxandybe the lengths of each leg of a right triangle, in cm.
hypotenuseh(x,y)=px2+y2 areaa(x,y)=1 2xy The maximum possible errors areh(3.05, 4.05)−h(3, 4) and
a(3.05, 4.05)−a(3, 4).
We estimate these using differentials.
dh=p x
x2+y2dx+ y
x2+y2dy ⇒dh=p 3
32+42(0.05)+ 4
p
32+42(0.05)=0.07cm
da=12ydx+ 1
2xdy ⇒da=
1
Remark.
The concept of differentials can also be extended to functions ofn
variables,n∈N. That is, ifuis a differentiable function ofx1,x2, . . .,xn−1 andxn, then
du= ∂u
∂xi
dxi+ ∂
u
∂x2
dx2+ · · · + ∂
u
∂xn
dxn,
Differentiability, Differentials and Local Linear
Approximation
Mathematics 54 - Elementary Analysis 2
Local Linear Approximation
Recall.Ifz=f(x,y) is differentiable at (x0,y0), then∆f ≈dz. That is
f(x0+∆x,y0+∆y)−f(x0,y0)≈fx(x0,y0)∆x+fy(x0,y0)∆y.
Letx=x0+∆xandy=y0+∆y. The above equation can be rewritten as
f(x,y)≈f(x0,y0)+fx(x0,y0)(x−x0)+fy(x0,y0)(y−y0).
Sincef is differentiable at (x0,y0), the error in approximation becomes smaller as (x,y) gets closer to (x0,y0).
Local Linear Approximation
Definition.
Letf(x,y) be differentiable at (x0,y0). Thelocal linear approximation off at (x0,y0) is defined as
L(x,y)=f(x0,y0)+fx(x0,y0)(x−x0)+fy(x0,y0)(y−y0).
Local Linear Approximation
Example.
Find the linearization approximation off(x,y)=exsinyat the origin.
Solution.Computing the partial derivatives, we have
fx(x,y)=exsiny
fx(0, 0)=0
fy(x,y)=excosy
fy(0, 0)=1
Thus, the linearization off(x,y) at (0, 0) is
L(x,y)=f(0, 0)+0(x−0)+1(y−0)+ =y.
That is,
f(x,y)≈y
Local Linear Approximation
Example
Approximatep 1
(3.92)2+(3.01)2using linear approximation.
Solution.First, we find a suitable function to be used, say
f(x,y)=p 1
x2+y2.
Since (3.92, 3.01) is ’close’ to (4, 3), the problem then reduces to approximatingf(3.92, 3.01).
To do this, we use the local linear approximation off at the point where (x,y)=(4, 3).
Computing the partial derivatives, we have
fx(x,y)= −
x
(x2+y2)3/2 fy(x,y)= −y
(x2+y2)3/2
fx(4, 3)= − 4
125 fy(4, 3)=
Local Linear Approximation
Example.(continued)
Approximate 1 p
(3.92)2+(3.01)2using linear approximation.
Solution.
Thus, the linear approximation off at the point (4, 3) is
f(4, 3)− 4
125(x−4)− 3
125(y−3).
That is,
f(x,y)=p 1
x2+y2≈ −4
125(x−4)+
−3
125(y−3)+ 1 5,
for all (x,y) close to (4, 3). Finally,
1 p
(3.92)2+(3.01)2=f(3.92, 3.01)≈ −4
125(3.92−4)+
−3
125(3.01−3)+ 1 5=
Local Linear Approximation
Remark.
