Linear Approximations and Differentials
Linear Approximation (or Tangent Line Approximation): To estimate
the value of a function (which might be difficult or impossible to evaluate
without electronic aids) near a certain known point, x =a, on its graph by
substituting its tangent line (because the values of a linear function is very
easy to compute) in its place. The price of such a simplification is that the
estimate is usually good only for a small interval around the given point, that
is, when x is close to a.
Hence, given a function f , and suppose (x0, y0) = (a, f (a)) is a known point
on the graph of f . Then the line tangent to the graph y = f (x) at a has slope
m = f ′(a). Therefore, the point-slope form of the equation of tangent line is
y − y0 = m(x − x0)

y − f (a) = f ′(a) (x − a)
Simplified to the slope-intercept form:
y = f (a) + f ′(a) (x − a)
That is, when x is near a , f (x) ≈ f (a) + f ′(a) (x − a).
The expression
L(x) = f (a) + f ′(a) (x − a)
is called the linearization of f at a.
Ex. Use linear approximation to estimate the value of sin(1).
First we need a known value of sin(x) near x = 1. The best point to use is at
the reference angle π/3 ≈ 1.047, since it is the angle closest to x = 1. Hence,
we will use the point (x0, y0) = (π/3, 3 /2).
Let f (x) = sin x, then f ′(x) = cos x.
The line tangent to y = sin x at x = π/3, therefore, has slope m = f ′(π/3) =
cos(π/3) = 1/2, and passes through the point (π/3, 3 /2). Its equation is,
therefore,
3 1

y
 x  
2 2
3
Simplifying,
 3 
1

3 1
y  x 
 x  
 
2
6
2
2
2
6

Hence, the linearization of y = sin x at x = π/3 is
L( x ) 
 3 
1
x  
 
2
2
6

At x = 1,
 3   1 1.732 3.1416
1
L(1)  (1)  
   

 0.5  0.866  0.5236  0.8424
2
2
6
2
2
6


The actual value of sin(1) is 0.84147…, so the estimate is accurate within
0.001.
Differential: To estimate the change in y-value of a function given the
change in x-value. Here, the change in x, ∆x = dx, is considered to be the
independent variable. The differential dy is the dependent variable and a
function of dx. The relation between dx and dy is given by
dy
 f (x)
dx
Therefore,
dy = f ′(x) dx
Here, the value of x is fixed at some number a, so it is actually a constant.
The independent variable is dx, and dy is the dependent variable.
Comment: The differential is just another way of looking at the linear
approximation. When using the differential, we estimate the change
in the y-value, as the x-value deviates from a fixed number a, rather
than the new y-value itself.
Notation: ∆x = dx is the change in x; ∆y = f (a + ∆x) − f (a) is the
change in y (the actual change of function’s values from x = a to x = a
+ ∆x); and the differential, dy = f ′(x) dx, is the estimated change of
function’s values from x = a to x = a + ∆x, according to the linear
approximation of f at a. In other words, dy = L(a + ∆x) − f (a), where
L(x) is the linearization of f at a.
To see the relationship between differential and linear approximation, recall
that the linearization of f at a is L(x) = f (a) + f ′(a) (x − a). When x is
moved from a to a + ∆x, the expression becomes
L(a + ∆x) = f (a) + f ′(a) ((a + ∆x) − a) = f (a) + f ′(a) ∆x
Since ∆x = dx, L(a + ∆x) = f (a) + f ′(a) dx and
dy = L(a + ∆x) − f (a) = (f (a) + f ′(a) dx) − f (a) = f ′(a) dx.
Therefore, for any fixed value x, and a variable value dx
dy = f ′(x) dx.
Comment: Since linear approximations work best for values close to the
fixed point a, so do differential become better estimates if the deviation dx
becomes smaller (that is, if x is close to a).
Ex. Compute ∆y and dy for y =
(i.) ∆x = 1.
(ii.) ∆x = 4.
f ( x) 
x for x = 1 and
1 1 / 2
1
x

2
2 x
At the fixed point x = a = 1, f ′(1) = 1/2.
(i.)
∆y = f (a + ∆x) − f (a) = f (2) − f (1) =
2 − 1 = 0.414213…
dy = f ′(1) dx = f ′(1) ∆x = (0.5)(1) = 0.5
Therefore, the linear approximation with base point a = 1 would give an
estimate of f (1) + dy = 1.5 for the value of 2 , a reasonable estimate
compares to the actual value of 1.414213…
(ii.)
∆y = f (a + ∆x) − f (a) = f (5) − f (1) = 5 − 1 = 1.236067…
dy = f ′(1) dx = f ′(1) ∆x = (0.5)(4) = 2
Therefore, the linear approximation with base point a = 1 would give an
estimate of f (1) + dy = 3 for the value of 5 (compares to the actual value
2.236067…), which is not too good an approximation. The dx in this case is
4, which is too large a deviation from the base point, x = 1, to give an
accurate result.