Math 400 (Section 4.5)
Exploration: Differentials & Linear Approximations
Name _______________________
1. Suppose that a function f is differentiable at c. Find an
equation for the line tangent to the graph of f at the
point ( c, f ( c ) ) . [Hint: Your equation will contain the
following symbols: y, f ( c ) , f ′ ( c ) , x and c . Use pointslope form for a line.]
Equation of line: _____________________________
Notice that in Problem 1, the function f and the point of tangency were completely arbitrary. This
was intentional, for now we have now exposed a general principle:
Linear Approximation to f at c
If a function f is differentiable on an interval I containing c, then the line
_______________________________ (for all x in I)
is the linear approximation to the function f at c.
Important: Since lim f ( x ) = f ( c ) , and ( c, f ( c ) ) is on the line, we can see that the linear
x →c
approximation is guaranteed to have very little error if x (in I) is very close to c.
2. Let f ( x ) = x 2 .
(a) Write the equation for the linear approximation to f at x = 1.
y = _________________
(b) Use the linear approximation to estimate f (1.2 ) .
f (1.2 ) ≈ _________
(c) Now find f ( 2 ) .
f ( x ) = _________
(d) The percent error of an approximation is defined by
approximate value - exact value
%Error =
exact value
Find the percent error for your approximation in (b).
% Error = _________
The figure at right shows the graph of y = f ( x ) ,
and the graph of
L ( x ) = f ( c ) + f ′ ( c )( x − c ) ,
the line tangent to the graph of f at the point
( c, f ( c ) ) . As mentioned on the previous page,
this line is an approximation for f (x) on an
interval containing c. Since x = c + ∆x , we see
that
∆x = __________.
We call this very small change in x the
differential dx. That is, dx = ∆x . Notice that
the change in f on the interval [ c, c + ∆x ] is ∆y = f ( x ) − f ( c ) . This small change in f is
approximated by the change in L on the same interval. This change in the linear approximation
for f is called the differential dy. We see that
dy = L ( x ) − L ( c ) .
3. Notice that the slope of the line L ( x ) is
L ( x) − L (c)
x−c
= f ′(c) .
Thus
dy = L ( x ) − L ( c ) = ________________ = ________________.
Notice that dy is an approximation for ∆y , and dy approaches ∆y as ∆x approaches zero. This is
consistent with the fact that
f ( x) − f (c)
dy
∆y
.
= f ′ ( c ) = lim
= lim
x
→
c
∆
x
→
0
dx x =c
x−c
∆x
Thus we see that
dy = f ′ ( c ) ⋅ dx
Since c was an arbitrary value of x (about which f is differentiable), we can replace c in this latest
expression with x, and get
dy = f ′ ( x ) dx
4. What happens when we divide both sides of this equation by dx? Is this consistent with
symbols we have used to represent the derivative?
5. The differential dy is an approximation for the change in y (that is, ∆y ).
(a) Approximate the change in the value of f ( x ) = x ln x on the interval [ 4, 4.2] .
(b) How close is this to the actual change in the value of f ( x ) on [ 4, 4.2] ?