MATH241 Lecture Notes - Lecture 11: Polynomial, Power Rule, Exponential Function

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MATH241 - Lecture 11 - Derivatives of a Polynomial *not on midterm*
Example
:
Find the derivative
A) (constant)(x)f=c
B) (x)f=x
C) (x)f=x2
D) (x)f=xn
Solution:
A) (x)f= lim
h → 0 h
f(x + h) − f(x)
= lim
h → 0 h
cc
= lim
h → 0 h
0
= lim
h → 0 0 = 0
B) lim
h → 0 h
f(x + h) − f(x)
= lim
h → 0 h
x + hx
= lim
h → 0 h
h
= 1= lim
h → 0 1
C) lim
h → 0 h
f(x + h) − f(x)
= lim
h → 0 h
(x + h) x
22
= lim
h → 0 h
x + 2xh + hx
222
= lim
h → 0 h
2xh + h2
x= lim
h → 0 2 + h x= 2
D) lim
h → 0 h
f(x + h) − f(x)
= lim
h → 0 h
(x + h) x
nn
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(using binomial theorem)= lim
h → 0 h
x + nx h + x h + . . . + nxh + hx
n n−1 2
n(n − 1) n−2 2n − 1 nn
= lim
h → 0 h
nx h + x h + . . . + nxh + h
n − 1 2
n(n − 1) n−2 2n−1 n
h x h . . xh= lim
h → 0 nxn−1 +2
n(n − 1) n−2 2+. +nn−2 +hn−1
=nxn−1
And in general, if then - This is called the power rule.(x)f=xn(x)f=nxn−1
If we know the derivatives of and , we can determine the following derivatives:(x)f(x)g
(x) (x) (x)F=f+g
(x)F= lim
h → 0 h
f(x + h) + g(x + h) − f(x) − g(x)
= lim
h → 0 h
f(x + h) − f(x)+ lim
h → 0 h
g(x + h) − g(x)
(x) (x)= f+g
Sum Rule:
(x) (x) (x)F=f+g
(constant times g(x))(x) g(x)G=c
(x)G= lim
h → 0 h
c g(x + h) − c g(x)
=clim
h → 0 h
g(x + h) − g(x)
g(x)= c
In summary:
(c)
d
dx = 0
f(x)
d
dx c f(x)
[ ] =c
(power rule)
d
dx (nx )
n−1
(x) (x)
d
dx f(x) (x)
[±g]=f±g
*Note: We can use the power rule for all real numbers
Example
:
Find the derivative
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Document Summary

Math241 - lecture 11 - derivatives of a polynomial *not on midterm* D) f f f f (x) (x) (x) (x) (constant) A) f(x + h) f(x) h c c h (x) = lim h 0 lim h 0. = lim h 0 f(x + h) f(x) x + h x h h h h h 0 h. 2 x + 2xh + h x. 2 + h x x= 2 f(x + h) f(x) h n (x + h) x n h n x + nx n 1 h + = lim h 0 x n(n 1) n 2 2 h + . + h x (using binomial theorem) n 1 nx h + x n(n 1) n 2 2 h + . + h nxn 1 + h n(n 1) n 2 2 + . h x. This is called the power rule. f (x) and (x)g. + g f(x + h) + g(x + h) f(x) g(x)

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