4.5: Linear Approximations and Differentials
Mysterious radiation has caused all calculators to stop
working. It just so happens that at this time, you need to
find out the square root of 5. The best you are going to be
able to do now is approximate it. But how?
Let’s start with the function y  x
know what
4 is
at x = 4 because we
When we make a graph of this process, you will
get a better sense of why we are doing this.
First of all, what is the tangent line at x = 4?
1
y  ( x  4)  2
4
Now graph both the function and its tangent line…

Use the tangent line to the function y  x at x = 4
to approximate 5
y
1
( x  4)  2
4
Notice that for
numbers close to
4, the tangent line
is very close to the
curve itself…
So lets try plugging
5 into the tangent
line equation and
see what we get…

Use the tangent line to the function y  x at x = 4
to approximate 5
y
1
( x  4)  2
4
Notice how close the point on the line is to the curve…
So let’s plug 5 into
the tangent line to
get…

(4, 2)

(5, 5 )
(5, 2.23607)
y
1
(5  4)  2  2.25
4
As it turns out…
So in this case, our approximation
will be a very good one as long as we
use a number close to 4.
5  2.23607

For any function f (x), the tangent is a close approximation
of the function for some small distance from the tangent
point.
y
f  x  f  a
Notice that at points very
close to a, the tangent line
and the curve are almost
the same.
0
xa
x

For any function f (x), the tangent is a close approximation
of the function for some small distance from the tangent
point.
f  x  f  a
(a, f(a)) 
(a+h, y(a+h))

(a+h, f(a+h))
Where y = mt(x – a) + f(a)
For any function f (x), the tangent is a close approximation
of the function for some small distance from the tangent
point.
f  x  f  a
(a, f(a)) 
(a+h, y(a+h))

(a+h, f(a+h))
We call the equation of this tangent
the linearization of the function at a
and it is denoted by the function
notation L(x).
Problems in the book will also refer to “the linearization of f at a”
y – y0 = m(x – x0)
We take the point-slope form…
Make a few notational
substitutions…
L( x)  f (a)  f (a)(x  a)
and…
f  x   L  x  is the standard linear approximation of f at a.
Remember: The linearization is just the equation of the tangent line.
The use of the term L(x) is to make it known that you are using the
tangent line to make a linear approximation of the function in
question. But we are still just talking about the tangent line.
Try a few simple
linearizations near x = 0:
sin x
L  x

x
cos x

f  x
1  x 
k
L(x)  f (0)  f (0)( x  0)
L(x)  sin(0)  cos(0)( x  0)
L(x)  01(x 0)

tan x
L(x)  f (a)  f (a)( x  a)
L( x)  x

The graph would look like this:
y
L(x)  x
y  sin( x)





x


Notice how close the graphs are to each other near x = 0
Try a few simple
linearizations near zero:
sin x
L  x

x
cos x

1
tan x
x
f  x
1  x 
k
L(x)  f (a)  f (a)( x  a)
L(x)  f (0)  f (0)( x  0)
L(x)  sin(0)  cos(0)( x  0)
L(x)  01(x 0)
L( x)  x
1  kx

Now try the other three.
We can also approximate
5 using another approach called
Differentials
Chemistry students should recognize y and x.
Remember that taking a derivative assumes that both x
and y go to 0. When we did this
y
x
dy
dx
For this method, we are going to allow dx to be a
small number but not 0.
We’ll start with our function y  x
5
Differentials
dy
dx
y
x
Note that as long as dx and dy are going to 0,
then we are getting more precise. But as soon
as we start making dx > 0, then we are going in
the other direction so whatever we get by this
method is an approximation of the change in y
For this method, we are going to allow dx to be a
small number but not 0.
We’ll start with our function y  x
dy
1

dx 2 x
Because dy and dx originally had
limits of 0, we aren’t supposed to
dy
think of
as a fraction. But in this
dx
case we are going to let dx > 0 so we
can treat dy like a fraction

dy 
1
2 x
dx
dx
In which dy will be a small change
in 
y which we will determine by
multiplying the derivative by the
small change in x (dx).
Let’s remember what we are trying to accomplish here.
We are going to use y  x starting at x = 4
because we know without a calculator that y = 2
5 ?
Since dy = approximate change in y,
dy 
dx
let’s set dx = 1
2 x
Why choose 1 in this case?
Because like linearization, we are going to use y  x to
approximate 5 from x = 4
And when we add 0.25 (the
change in y) to 2 (our original y),
1
dy 
 1  0.25 we get the same approximation of 5
2 4
that we had with linearization.
1
y  dy  2  0.25  2.25
Use linearization and differentials to approximate
3
9
using the function y  3 x
dy
1
 3 2
dx 3 x
y x
3
Since
23 8
We’ll use the tangent line at x = 8 for the linearization

1
( x  8)  2
L(x) 
12
1
L(9)  2  2.083
12
3
9  2.080083823
Use linearization and differentials to approximate
3
9
using the function y  3 x
dy
1
 3 2
dx 3 x
y x
3
Now let’s use differentials:
Since
dy 
1
3 3 x 2
23 8
dx
Again, dx
  1 simply because it is the difference between 8
and 9
1
1 
 0.083
dy  3 2  1 
12
3 8
3
y  dy  2  .083  2.083
Same as with linearization
9  2.080083823
A square with sides x has an area
x

A( x)  x 2
If a 2 X 2 square has it’s sides
increase by 0.1, use differentials
to approximate how much its area
will increase.
x
x2
dA
 2x
dx
dx  0? .1
dA  2 x dx
dA  2(2)(0.1)  0.4
x
dx
 dA
What will the actual difference in area be?
x
A(2)  4
A(2.1)  4.41
Actual change in area  0.41
x
So our approximation is off by
0.41  0.4  0.01
Our percentage error here is:
Error
0.01

 2.44%
Actual Change 0.41
x
dx