Section 23 Basic Differentiation Formulas2010 Kiryl

Section 2.3 Basic Differentiation Formulas

2010 Kiryl Tsishchanka

Basic Differentiation Formulas
DERIVATIVE OF A CONSTANT FUNCTION: d (c) = 0 dx c′ = 0

or

Proof: Suppose f (x) = c, then f (x + h) − f (x) c−c 0
= lim
= lim = lim 0 = 0 h→0 h→0 h→0 h h→0 h h f ′ (x) = lim
EXAMPLES:
′

1 = 0,

′

5 = 0,

′

0 = 0,

′

(−7/9) = 0,

√
1+ 5
2

′

π = 0,

′

= 0,

√
(x 3 x + 1 − x4/3 )′ = 0

THE POWER RULE: If n is a positive integer, then d n
(x ) = nxn−1 dx or

(xn )′ = nxn−1

Proof: Before we prove this result rigorously, let us consider cases n = 2, 3, 4.
If n = 2, then x 2 − a2
(x − a)(x + a) f (x) − f (a)
= lim
= lim
= lim (x + a) = a + a = 2a x→a x − a x→a x→a x→a x−a x−a f ′ (a) = lim
If n = 3, then

x 3 − a3
(x − a)(x2 + xa + a2 ) f (x) − f (a)
= lim
= lim x→a x − a x→a x→a x−a x−a

f ′ (a) = lim

= lim (x2 + xa + a2 ) = a2 + a · a + a2 = 3a2 x→a If n = 4, then f (x) − f (a) x 4 − a4
(x − a)(x3 + x2 a + xa2 + a3 )
= lim
= lim x→a x→a x − a x→a x−a x−a f ′ (a) = lim

= lim (x3 + x2 a + xa2 + a3 ) = a3 + a2 · a + a · a2 + a3 = 4a3 x→a In general, we have f (x) − f (a) x n − an
(x − a)(xn−1 + xn−2 a + . . . + xan−2 + an−1 )
= lim
= lim x→a x→a x − a x→a x−a x−a f ′ (a) = lim

= lim (xn−1 + xn−2 a + . . . + xan−2 + an−1 ) = nan−1 x→a 1

Section 2.3 Basic Differentiation Formulas

2010 Kiryl Tsishchanka

EXAMPLES:
(a) If f (x) = x2 , then f ′ (x) = (x2 )′ = [n = 2] = 2x2−1 = 2x1 = 2x.
(b) If f (x) = x9 , then f ′ (x) = (x9 )′ = [n = 9] = 9x9−1 = 9x8 .
(c) If f (x) = x, then f ′ (x) = (x1 )′ = [n = 1] = 1 · x1−1 = 1 · x0 = 1 · 1 = 1.
THE POWER RULE (GENERAL VERSION): If n is any real number, then d n
(x ) = nxn−1 dx (xn )′ = nxn−1

or

EXAMPLES:
(a) If f (x) = x−4 , then f ′ (x) = (x−4 )′ = [n = −4] = (−4)x−4−1 = −4x−5 .
1
1
(b) If f (x) = , then f ′ (x) = (x−1 )′ = [n = −1] = (−1)x−1−1 = −x−2 = − 2 . x x
√
1
1
1
(c) If f (x) = x, then f ′ (x) = (x1/2 )′ = [n = 1/2] = x1/2−1 = x−1/2 = √ .
2
2
2 x
√
(d) If f (x) = x2 3 x, then
7
f ′ (x) = (x2 · x1/3 )′ =