17. Constant multiple rule, Sum rule
Constant multiple rule, Sum rule
Constant multiple rule
The rules we obtain for finding derivatives are of two types:
Sum rule
ˆ Rules for the derivatives of the basic functions, such as xn , cos x, sin x, ex , and so
forth. (We have already seen the rule for the first of these.)
ˆ Rules for how to find the derivative of a function built up of simpler functions that
we already know the derivatives of. For example, a rule that tells us how to find the
derivative of ex cos x once we know the derivatives of ex and cos x (product rule).
The two rules we get in this section, the constant multiple rule and the sum rule, are of
this second type.
17.1. Constant multiple rule
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Constant multiple rule. For any function f and any constant c,
d
d
[cf (x)] = c [f (x)] .
dx
dx
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In words, the derivative of a constant times f (x) equals the constant times the derivative
of f (x). Put another way, constant multiples slip outside the differentiation process. For
instance,
d 5
d 5
3x = 3
x = 3(5x4 ) = 15x4 .
dx
dx
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With g(x) = cf (x), the rule says that g 0 (x) = cf 0 (x), so we verify the rule by showing that
this equation holds:
Constant multiple rule, Sum rule
Constant multiple rule
g(x + h) − g(x)
g (x) = lim
h→0
h
cf (x + h) − cf (x)
= lim
h→0
h
f (x + h) − f (x)
= lim c ·
h→0
h
f (x + h) − f (x)
= c lim
h→0
h
0
= cf (x).
Sum rule
0
(by limit law (iii))
Using the constant multiple rule, we can show that the derivative of any constant function
is 0:
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Derivative of constant.
d
[c] = 0
dx
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(c, constant).
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d
d
[c] = c [1] = c(0) = 0 (see 16.3). This formula says that the
dx
dx
constant function f (x) = c (horizontal line) has general slope function f 0 (x) = 0, as we
might have guessed.
The reason is that
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17.2. Sum rule
Constant multiple rule, Sum rule
Constant multiple rule
Sum rule. For any functions f and g,
Sum rule
d
d
d
[f (x) + g(x)] =
[f (x)] +
[g(x)] .
dx
dx
dx
In words, the derivative of a sum is the sum of the derivatives. For instance,
d 3
d 6
d 3
x + x6 =
x +
x = 3x2 + 6x5 .
dx
dx
dx
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The verification of the sum rule is left to the exercises (see Exercise 17 – 2).
Since f (x) − g(x) can be written f (x) + (−1)g(x), it follows immediately from the sum rule
and the constant multiple rule that the derivative of a difference is the difference of the
derivatives:
d
d
d
[f (x) − g(x)] =
[f (x)] −
[g(x)] .
dx
dx
dx
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17.2.1
Solution
Example
d 5
2x + 4x3 − 7x2 + 6x + 9 .
Evaluate
dx
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Using, in turn, the sum rule, the constant multiple rule, and the power rule, we
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have
Constant multiple rule, Sum rule
d 5
2x + 4x3 −7x2 + 6x + 9
dx
d 5
d 3
d 2
d
d
=
2x +
4x −
7x +
[6x] +
[9]
dx
dx
dx
dx
dx
d 5
d 3
d 2
d
=2
x +4
x −7
x + 6 [x] + 0
dx
dx
dx
dx
= 2(5x4 ) + 4(3x2 ) − 7(2x1 ) + 6(1)
Constant multiple rule
Sum rule
= 10x4 + 12x2 − 14x + 6.
(We have also used that
is 0.)
d
[x] = 1 (see 16.3) and that the derivative of a constant function
dx
The preceding example shows that the derivative of a polynomial can be computed one
term at a time by bringing down the power of x and multiplying it by the coefficient, and
then reducing the power of x by 1. This makes finding the derivative of a polynomial a
one-step process. For example,
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d 4
3x − 5x3 + 8x2 + 2x − 7 = 12x3 − 15x2 + 16x + 2.
dx
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(Actually, instead of thinking about powers for the last two terms, we just use that the
derivative of a multiple of x is the multiple, and the derivative of a constant is 0.)
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17.2.2 Example
where x = −1.
Solution
Find an equation of the line tangent to the graph of f (x) = x4 − 4x2
We will use the point-slope form of the line, y − y0 = m(x − x0 ), which requires
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that we know the slope m of the line and a point (x0 , y0 ) on the line. The point can be
taken to be the point of tangency:
Constant multiple rule, Sum rule
Constant multiple rule
(x0 , y0 ) = (−1, f (−1)) = (−1, −3).
Sum rule
0
3
The slope m is the derivative of f evaluated at −1. Since f (x) = 4x − 8x, we have
m = f 0 (−1) = 4. The answer is y − (−3) = 4(x − (−1)), which simplifies to y = 4x + 1.
Here is the graph:
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Since the power rule holds even if the power of x is not a positive integer, the procedure
just described is valid not only for polynomials, but also for expressions such as the one in
the following example.
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17.2.3
Example
√
5
2
Find the derivative of f (x) = 4 x + 3 − √
.
x
( 3 x)2
Constant multiple rule, Sum rule
Constant multiple rule
Solution
We first rewrite the expression to get powers of x:
Sum rule
f (x) = 4x1/2 + 5x−3 − 2x−2/3 .
Now we compute the derivative and finish by writing the answer using the same notation
as in the statement of the problem:
i
d h 1/2
4x + 5x−3 − 2x−2/3
dx
= 2x−1/2 − 15x−4 + 43 x−5/3
15
4
2
=√ − 4+ √
.
x x
3( 3 x)5
f 0 (x) =
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17 – Exercises
Constant multiple rule, Sum rule
Constant multiple rule
Sum rule
17 – 1
Find the derivatives of the following functions:
(a) f (x) = 4x5 − 8x3 + x2 − x + 7
√
4
(b) f (t) = 3 t + √
3
4
t
x − 2x +
(c) f (x) =
x
17 – 2
√
x−3
(Hint: First rewrite.)
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Verify the sum rule (see 17.2).
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Hint: Let s(x) = f (x) + g(x). The sum rule says that s0 (x) = f 0 (x) + g 0 (x). Use the
definition of the derivative (as applied to the function s) to show that this equation holds.
(It might help to study the verification of the constant multiple rule.)
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