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Sullivan 04 apcalc4e 45342 ch02 166 233 5pp August 7, 2023 12:54

196 Chapter 2 • The Derivative and Its Properties

THEOREM Difference Rule
If the functions f and g are differentiable and if F(x) = f (x) − g(x), then F is
′
′
differentiable, and F (x) = f (x) − g (x). That is,
′
IN WORDS The derivative of the difference
d d d
of two differentiable functions is the [ f (x) − g(x)] = f (x) − g(x)
dx dx dx
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difference of their derivatives. That is,
( f − g) = f − g . ′
′
′
The proof of the Difference Rule is left as an exercise. (See Problem 78.)
The Sum and Difference Rules extend to sums (or differences) of more
than two functions. That is, if the functions f 1 , f 2 , . . . , f n are all differentiable,
and a 1 , a 2 , . . . , a n are constants, then
d d d d
[a 1 f 1 (x) + a 2 f 2 (x) + · · · + a n f n (x)] = a 1 f 1 (x) + a 2 f 2 (x) + · · · + a n f n (x)
dx dx dx dx

Combining the rules for finding the derivative of a constant, a power function, and
a sum or difference allows us to differentiate any polynomial function.
EXAMPLE 5 Differentiating a Polynomial Function

2
4
(a) Find the derivative of f (x) = 2x − 6x + 2x − 3.
′
(b) What is f (2)?
(c) Find the slope of the tangent line to the graph of f at the point (1, −5).
(d) Find an equation of the tangent line to the graph of f at the point (1, −5).
(e) Find an equation of the normal line to the graph of f at the point (1, −5).
(f) Use technology to graph f , the tangent line, and the normal line to the graph of f
at the point (1, −5) on the same screen.

Solution
d 4 2 d 4 d 2 d d
(a) f (x) = (2x − 6x + 2x − 3) = (2x ) − (6x ) + (2x) − 3
′
dx ↑ dx dx dx dx
Sum/Difference Rules
d 4 d 2 d
= 2 · x − 6 · x + 2 · x − 0
dx dx dx
↑
Constant Multiple Rule
3
3
= 2 · 4x − 6 · 2x + 2 · 1 = 8x − 12x + 2
↑ ↑
Simple Power Rule Simplify
3
(b) f (2) = 8 · 2 − 12 · 2 + 2 = 64 − 24 + 2 = 42.
′
(c) The slope of the tangent line to the graph of f at the point (1, − 5) equals f (1).
′
3
f (1) = 8 · 1 − 12 · 1 + 2 = 8 − 12 + 2 = −2
′
(d) Use the point-slope form of an equation of a line to find an equation of the tangent
line to the graph of f at (1, −5).
y − (−5) = −2(x − 1)
y = −2(x − 1) −5 = −2x + 2 − 5 = −2x − 3
4
2
The line y = −2x − 3 is tangent to the graph of f (x) = 2x − 6x + 2x − 3 at the
point (1, −5).
(e) Since the normal line and the tangent line at the point (1, −5) on the graph of f
are perpendicular and the slope of the tangent line is −2, the slope of the normal
1
line is .
2
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