2.5 Derivatives of Polynomials

1

The four-step process has been introduced to found derivatives.In the present and the next lecture, using this process, we develop differentiation rules to directly apply in the future.

Constant Rule:If c is a constant, then 0

dx

dc

Proof:The given function is y=c. We apply the four-step process to this function: Step 1. y=f(x)=c, y+y=f(xxc. Step 2. y=c - c=0.

Step 3. y/x = 0. Step 4.

Picture: the function is a horizontal line, and its slope at any point is 0.

.0lim0

x

yx

2

y=x is a diagonal line. So, its slope is 1: dx/dx = 1. Prove!

1 nn

nxdx

dxPower Rule:If n>0,

Proof (for integer n>0 only):Again, we apply the four-step process to this function:Step 1. Step 2.

Step 3.

Step 4. .lim 1

0

n

xnx

x

y

...)( xxfyy

.)(

)(

)(

1

1

nn

nnnn

nn

xxnx

xxxnxx

xxxy

.)( 11 nn xnx

x

y

3

dx

dv

dx

duvu

dx

d )(

Sum Rule: For any functions of x, u(x) and v(x),

Proof:Again, we apply the four-step process to the sum:Step 1. Denote u=f(x), v=g(x), y=h(x)=u+v=f(x)+g(x), then y+y = h(xxf(x+x)+g(x+x).Step 2. y = h(x+x)-h(x) = f(x+x)+g(x+x)-f(x)-g(x). Denote u=f(x+x)-f(x), v=g(x+x)-g(x). Then, y=u+vStep 3.

Step 4. Using the theorem that the limit of a sum is equal to the sum of limits (Sec. 2.2, p. 64), we state

.limlim][limlim0000 dx

dv

dx

du

x

v

x

u

x

v

x

u

x

y

dx

dyxxxx

.x

v

x

u

x

y

4

dx

duccu

dx

d

Constant Multiplier Rule: For any constant c and function u(x),

Prove the above statement doing steps 1 through 4. At the last step, you need to apply another part of the theorem from Sec. 2.2 (p.64):

For any constant c, ).(lim)(lim xfcxcfaxax

Example: Differentiate 132 23 xxy

)1(6)2(3)3(2}Power{

)(3)(2}Mult. Const.{)3()2(

} Const.{)1()3()2(}Sum{

1213

2323

23

xxxx

xdx

dx

dx

dx

dx

dx

dx

ddx

dx

dx

dx

dx

d

dx

dy

5

Exercises: Differentiate

23

1 25 tty

222 y

xe

exy22

xxy 5.1

6

HomeworkSection 2.5: 3,9,13,17,19,21,23,27.