131notesSection 3.9 Differentials and Linear Approximations
Recall that the tangent line to a differentiable function f at (a, f(a)) is a good
approximation to f(x) if x is close enough to a.
y  f ' ( a )( x  a )  f ( a )
The tangent line is
y  f ( a )  f ' ( a )( x  a )
If we write  variable for the exact change in the variable then x-a =
f(x) - f(a) =  f

x and
Since f(x) is close to y= the value on the tangent line at x , we can write

Definition:
 y  f (a )  f ' (a )x
f
df  f ' ( a )  x
=f '(a)dx where

x=dx. df is the differential of f near a.
Since f ' is a function and we can do this at any a, we also write df = f '(x)dx
1
Example:
f (x)  x
Approximate
3
df 
1

2
x 3 dx
3
3
9
=f (9).
We know f(8) = 2 so let a = 8. Then dx = 9-8 = 1 and f '(8)=
Then
3
9 2
1
12
so df =
1
12
1
and 3 9  2 
12
1
12
Example: Approximate tan x near 0.
f ( x )  tan x
a 0
f ' ( x )  sec
2
f ' (0)  1
x
tan0 = 0 and  x=x-0=x y = x is the tangent line to f at 0.
tan x  x if x is near 0.
Example: A snowball melts and maintains its spherical shape. Approximate the change in
the volume if the radius decreases by 2 cm.
For r = the radius in cm.
V 
4
 r
3
V ' ( r )  4 r
2
dr   2
3
2
dV  4  r dr   8 r
2
cubiccm .
Example: Approximate the % change in the volume of a cube if the sides increase by 1%.
Let x be the side of the cube. 1% of x = .01x = dx
V  x
dV
V
3
 . 03
2
2
dV  3 x dx  3 x (. 01 x )  . 03 x
a change of 3 %
3