for any function f (x), the tangent is a close approximation of the function for some small distance...
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For any function f (x), the tangent is a close approximation of the function for some small distance from the tangent point.
y
x0 x a
f x f aWe call the equation of the tangent the linearization of the function.
The linearization is the equation of the tangent line, and you can use the old formulas if you like.
Start with the point/slope equation:
1 1y y m x x 1x a 1y f a m f a
y f a f a x a
y f a f a x a
L x f a f a x a linearization of f at a
f x L x is the standard linear approximation of f at a.
Linearization
Example Finding a Linearization
Find the linearization of ( ) cos at / 2 and use it to approximate
cos 1.75 without a calculator.
f x x x
Since ( / 2) cos( / 2) 0, the point of tangency is ( / 2,0). The slope of the
tangent line is '( / 2) sin( / 2) 1. Thus ( ) 0 ( 1) .2 2
To approximate cos 1.75 (1.75) (1.75) 1.75 .2
f
f L x x x
f L
Important linearizations for x near zero:
1k
x 1 kx
sin x
cos x
tan x
x
1
x
1
21
1 1 12
x x x
13 4 4 3
4 4
1 5 1 5
1 51 5 1
3 3
x x
x x
f x L x
This formula also leads to non-linear approximations:
Differentials:
When we first started to talk about derivatives, we said that
becomes when the change in x and change in
y become very small.
y
x
dy
dx
dy can be considered a very small change in y.
dx can be considered a very small change in x.
Estimating Change with Differentials
Let be a differentiable function.
The differential is an independent variable.
The differential is:
y f xdxdy dy f x dx
Example Finding the Differential dy
5
Find the differential and evaluate for the given value of and .
2 , 1, 0.01
dy dy x dx
y x x x dx
45 2
5 2 0.01
0.07
dy x dx
dy
Examples
Find dy if
a. y = x5 + 37x
Ans: dy = (5x4 + 37) dx
b. y = sin 3x
Ans: dy = (3 cos 3x) dx
Differential Estimate of Change
Let f(x) be differentiable at x = a. The approximate change in the value of f when x changes from a to a + dx is
df = f ‘(a) dx.
Example
The radius r of a circle increases from a = 10 m to 10.1 m. Use dA to estimate the increase in circle’s area A.
Compare this to the true change ΔA.
Example: Consider a circle of radius 10. If the radius increases by 0.1, approximately how much will the area change?
2A r
2 dA r dr
2 dA dr
rdx dx
very small change in A
very small change in r
2 10 0.1dA
2dA (approximate change in area)
2dA (approximate change in area)
Compare to actual change:
New area:
Old area:
210.1 102.01
210 100.00
2.01
.01
2.01
Error
Actual Answer.0049751 0.5%