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Section 2.3 The Power Rule
Let’s first find some common derivatives
Theorem For any constant c, d dx
(c) = 0
Proof Let f (x) = c
d dx (c) = lim h→ 0
f (x + h) − f (x) h = lim h→ 0
c − c h = lim h→ 0 0 = 0
Theorem Let f (x) = x, then d dx
(x) = 1
Proof
f ′(x) = lim h→ 0 f^ (x^ +^ h h)^ −^ f^ (x)
= lim h→ 0 (x^ +^ hh) −^ x= lim h→ 0 1 = 1
Theorem (The Power Rule) For any integer n > 0, if f (x) = xn, then
d dx
(xn) = nxn−^1
Proof The proof of this theorem requires us to recall the Binomial Theorem from PreCalc:
(x+h)n^ = xn^ +nxn−^1 h+ n(n^ −^ 1) 2
xn−^2 h^2 +...+
n k
xn−khk^ +...nxhn−^1 +hn
where
(n k
= n(n−1) 1 · 2 ...· 3 (...kn−k+1)
d dx (x
n) = lim h→ 0
f (x + h) − f (x) h = lim h→ 0
(x + h)n^ − xn h
= lim h→ 0
xn^ + nxn−^1 h + n(n 2 − 1)xn−^2 h^2 + ... +
(n k
xn−khk^ + ...nxhn−^1 + hn^ − xn h
= lim h→ 0
nxn−^1 h + n(n 2 − 1)xn−^2 h^2 + ... +
(n k
xn−khk^ + ...nxhn−^1 + hn h = lim h→ 0 nxn−^1 +
n(n − 1) 2 x
n− (^2) h + ... +
n k
xn−khk−^1 + ...nxhn−^1 + hn−^1 ︸ ︷︷ ︸ there is an h in every term here = nxn−^1 + 0 + 0 + .... + 0 = nxn−^1
Example Let f (x) = x^2008 then by Power Rule f ′(x) = 2008x^2008 −^1 = 2008 x^2007 The above theorem is true for integers. What if the power is a fraction? Let’s check on an example;
Example Let f (x) = √x and find f ′(x)?
f ′(x) = lim h→ 0
x + h −
x h = lim h→ 0
x + h −
x h ·
x + h +
√ x x + h +
x = lim h→ 0
h h(
x + h +
x)
x =
2 x
− 1 / 2
So f (x) = x^1 /^2 gives f ′(x) = 12 x−^1 /^2 = 12 x^1 /^2 −^1 which is like the Power Rule.
In fact the Power Rule holds for any real number as we will state in the next theorem. For the proof of this theorem you’ll need to wait until ”logarith- mic” differentiation though.
Theorem (The General Power Rule) For any real number r,
d dx (x
r) = rxr− 1
Once again the last equality is due the fact that we assumed f to be differ- entiable.
Example f (x) = 4x^5 + 3x^2
x − (^1) x find f ′(x).
First write the function as the sum of power functions after all those are the only ones so far we know how to differentiate. f (x) = 4x^5 + 3x^5 /^2 − x−^1
f ′(x) = (x^4 )′^ + (3x^5 /^2 )′^ − (x−^1 )′^ by General Derivative Rules 1 = (x^4 )′^ + 3(x^5 /^2 )′^ − (x−^1 )′^ by General Derivative Rules 2 = 4 · 5 x^4 + 3 · 5 2
x^3 /^2 − (−1)x−^2
= 20x^4 +^15 2
x^3 /^2 +^1 x^2 Example f (x) = x^2 +4√xx+3 Find f ′(x) Once again first write the function as the sum of power functions first; f (x) = √x^2 x + √^4 xx + √^3 x = x^3 /^2 + 4x^1 /^2 + 3x−^1 /^2
f ′(x) = (x^3 /^2 )′^ + (4x^1 /^2 )′^ + 3(x−^1 /^2 )′^ by General Derivative Rules 1 = (x^3 /^2 )′^ + 4(x^1 /^2 )′^ + 3(x−^1 /^2 )′^ by General Derivative Rules 2 =^3 2
x^1 /^2 + 2x−^1 /^2 − 3 2
x−^3 /^2
=^3 2
x + √^2 x
2 x^3 /^2 Higher Derivatives
If f is a differentiable function, then its derivative f ′^ is also a function, so f ′ may have a derivative of its own.
Example Let f (x) = x^3 − 6 x^2 − 5 x + 3. By derivative rules and the power rule we know: df dx = 3x^2 −^12 x^ −^5 d^2 f dx^2 = 6x^ −^12 d^3 f dx d (^4) f^3 = 6, dx^4 = 0
Note that ddxnfn = 0 for all n ≥ 4
