1 Section 2.3 The

Let’s first find some common

d Theorem For any constant c, (c) = 0 dx

Proof Let f(x) = c

d f(x + h) − f(x) (c) = lim dx h→0 h c − c = lim = lim 0 = 0 h→0 h h→0

d Theorem Let f(x) = x, then (x) = 1 dx

Proof

f(x + h) − f(x) f 0(x) = lim h→0 h (x + h) − x = lim = lim 1 = 1 h→0 h h→0

Theorem (The Power Rule) For any n > 0, if f(x) = xn, then

d (xn) = nxn−1 dx

Proof The proof of this theorem requires us to recall the Theorem from PreCalc:

µ ¶ n(n − 1) n (x+h)n = xn +nxn−1h+ xn−2h2 +...+ xn−khk +...nxhn−1 +hn 2 k

¡n¢ n(n−1)...(n−k+1) where k = 1·2·3...k 2

d f(x + h) − f(x) (xn) = lim dx h→0 h (x + h)n − xn = lim h→0 h ¡ ¢ xn + nxn−1h + n(n−1) xn−2h2 + ... + n xn−khk + ...nxhn−1 + hn − xn = lim 2 k h→0 h ¡ ¢ nxn−1h + n(n−1) xn−2h2 + ... + n xn−khk + ...nxhn−1 + hn = lim 2 k h→0 h µ ¶ n(n − 1) n = lim nxn−1 + xn−2h + ... + xn−khk−1 + ...nxhn−1 + hn−1 h→0 2 k | {z } there is an h in every term here = nxn−1 + 0 + 0 + .... + 0 = nxn−1 Example Let f(x) = x2008 then by Power Rule f 0(x) = 2008x2008−1 = 2008x2007 The above theorem is true for . What if the power is a ? Let’s check on an example; √ Example Let f(x) = x and find f 0(x)?

√ √ x + h − x f 0(x) = lim h→0 h √ √ √ √ x + h − x x + h + x = lim · √ √ h→0 h x + h + x h 1 1 = lim √ √ = √ = x−1/2 h→0 h( x + h + x) 2 x 2 1/2 0 1 −1/2 1 1/2−1 So f(x) = x gives f (x) = 2 x = 2 x 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 (xr) = rxr−1 dx 3 √ Example For f(x) = 3 x4 = x4/3, find f 0(x). 4 0 4 3 −1 4 1/3 f (x) = 3 x = 3 x

Example For g(x) = xπ ⇒ g0(x) = πxπ−1

Theorem (General Rules) If f(x) and g(x) are differentiable at x, and c is any constant then,

d d d 1) dx [f(x) ± g(x)] = dx [f(x)] + dx [g(x)]

d d 2) dx [cf(x)] = c dx [f(x)]

Proof 1) Let y = f(x) + g(x)

[f(x + h) + g(x + h)] − [f(x) + g(x)] y0 = lim h→0 h [f(x + h) − f(x)] + [g(x + h) − g(x)] = lim h→0 h [f(x + h) − f(x)] [g(x + h) − g(x)] = lim + lim h→0 h h→0 h d d = [f(x)] + [g(x)] dx dx The last equality above is due the fact that both individual limits exists because we assumed f and g to be differentiable. You can prove the y = f(x) − g(x) case the same way I did above.

2) Let y = cf(x).

cf(x + h) − cf(x) y0 = lim h→0 · h ¸ f(x + h) − f(x) = lim c h→0 h · ¸ f(x + h) − f(x) = c lim by Law 2 h→0 h d = c [f(x)] dx 4

Once again the last equality is due the fact that we assumed f to be differ- entiable.

5 2√ 1 0 Example f(x) = 4x + 3x x − 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) = 4x5 + 3x5/2 − x−1

f 0(x) = (x4)0 + (3x5/2)0 − (x−1)0 by General Derivative Rules 1 = (x4)0 + 3(x5/2)0 − (x−1)0 by General Derivative Rules 2 5 = 4 · 5x4 + 3 · x3/2 − (−1)x−2 2 15 1 = 20x4 + x3/2 + 2 x2 2 x +4√x+3 0 Example f(x) = x Find f (x) Once again first write the function as the sum of power functions first; 2 √x √4x √3 3/2 1/2 −1/2 f(x) = x + x + x = x + 4x + 3x f 0(x) = (x3/2)0 + (4x1/2)0 + 3(x−1/2)0 by General Derivative Rules 1 = (x3/2)0 + 4(x1/2)0 + 3(x−1/2)0 by General Derivative Rules 2 3 3 = x1/2 + 2x−1/2 − x−3/2 2 2 3√ 2 3 = x + √ − 2 x 2x3/2 Higher Derivatives

If f is a differentiable function, then its derivative f 0 is also a function, so f 0 may have a derivative of its own.

Example Let f(x) = x3 − 6x2 − 5x + 3. By derivative rules and the power rule we know: df 2 dx = 3x − 12x − 5 d2f dx2 = 6x − 12 d3f dx3 = 6, d4f dx4 = 0

dnf Note that dxn = 0 for all n ≥ 4