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MIT 单变量微积分lecture note+problem set+exam.pdf
18.01 Single Variable Calculus
Fall 2006
Lecture 1
18.01 Fall 2006
Unit 1: Derivatives
A. What is a derivative?
• G e o m e t r i c interpretation
• Ph}^sical interpretation
參 I m p o r t a n t for an}，m e a s u r e m e n t (economics, political science, finance, plxysics, etc.)
B. How to differentiate any function you know.
• For example:
d x 、
differentiate an}^ function is the subject of the first t w o w e e k s of this course.
arctana;^ \ye wn\ discuss w h a t a derivative is toda}^. Figuring out h o w to
■
Lecture 1: Derivatives, Slope, Velocity, and Rate
of Change
Geometric Viewpoint on Derivatives
Figure 1 ： A function with secant and tangent lines
T h e derivative is the slope of the line tangent to the g r a p h of /(x). B u t w h a t is a tangent line,
exactl}^?
1
Lecture 1
18.01 Fall 2006
• It is N O T just a line that m e e t s the g r a p h at o n e point.
• It is the limit of the secant line (a line d r a w n b e t w e e n t w o points o n the graph) as the distance
b e t w e e n the t w o points goes to zero.
Geometric definition of the derivative:
Limit of slopes of secant lines P Q as Q ^ P ( P fixed). T h e slope of PQ:
Q
( x 〇+ A x , f ( x 〇+ A x ) )
Secant Line
limAx^O A x
lim / ( x 〇 十 A ：r) — / ( x 〇)
A x
“difference quotient”
f ( x 〇)
^derivative of / at x 〇r,
Example 1. f(x)
1
x
O n e thing to keep in m i n d w h e n w o r k i n g with derivatives: it m a y b e t e m p t i n g to plug in A x = 0
right away. If y o u d o this, however, y o u will always e n d u p with —— = 。 Y o u will always n e e d to
d o s o m e cancellation to get at the answer.
A x
0
A /
A x
l
A x
1
A x
x 〇 — (x〇 十 A x )
(x〇 十 A x ) x 〇
1
A x
—A x
- 1
(x〇 十 A x ) x 〇
(x〇 十 A x ) x 〇
T a k i n g the limit as A x ^ 0,
lim. ------ ---：— = —9
Xq
A x —0 (X〇 十 A x ) X 〇
2
Lecture 1
18.01 Fall 2006
Hence,
f ' ( x 〇)
Notice that / ;(x〇) is negative —— as is the slope of the tangent line o n the g r a p h above.
Finding the tangent line.
Wri te the equation for the tangent line at the point (x〇,y〇) using the equation for a line, w h i c h }^ou
all learned in high school algebra:
y - y 〇 = ff(x〇)(x-x 〇)
P l u g in y 〇 = / ( x 〇) = — a n d / ;(x〇) = — to get:
'
1
X 〇
一 1
X g
Lecture 1
18.01 Fall 2006
Just for fun, \eVs c o m p u t e the area of the triangle that the tangent line forms wi th the x- a n d
3^-axes (see the s h a d e d region in Fig. 4).
First calculate the x-intercept of this tangent line. T h e x-intercept is w h e r e y = 0. P l u g y = 0
into the equation for this tangent line to get:
1
x 〇
- 1
x 〇
1
_ 2"*^
x 0
=
=
r
- 1 .
x
0
- 1
- 2 X
x
0
2
---
x 〇
- X 〇)
~
1
x 〇
X
=
工 5 ( 二 ） = : 2
X〇
So, the x-intercept of this tangent line is at x = 2 x 〇.
N e x t w e claim that the 3^-intercept is at y = 2y〇. Since y = — a n d x = — are identical equations,
the g r a p h is s}^mmetric w h e n x a n d y are exchanged.
S3^mmetr}^ then, the 3^-intercept is at
y = 2 y 〇. If }，o u d o n ’t trust reasoning with S3，m m e t r } ，，}，o u m a } ，follow the s a m e chain of algebraic
reasoning that w e used in finding the x-intercept. ( R e m e m b e r , the 3^-intercept is w h e r e x = 0.)
x
y
Finall}，，
A r e a = i(2}，0)(2x0) = 2x 〇3,〇 = 2 x 〇(丄）= 2 (see Fig. 5)
1
x 〇
Curiousl}^, the area of the triangle is ahuays 2, n o m atter w h e r e o n the g r a p h w e d r a w the tangent
line.
4
Lecture 1
18.01 Fall 2006
Notations
Calculus, rather like English or an}^ other language, w a s developed
several people. A s a result,
just as there are man}^ wa}^s to express the s a m e thing, there are man}^ notations for the derivative.
Since y = /( ；r), it7s natural to write
Ay = A / = / 〇 ) - / 〇〇) = /〇〇 十 Ax) - / 〇〇)
W e sa)，“Delta y ” or “Delta / ” or the “c h a n g e in y ”.
If w e divide b o t h sides
A x = x — x 〇, w e get t w o expressions for the difference quotient:
T a k i n g the limit as A x ^ 0, w e get
A ^ = A /
A x
A x
A y
A x
A / 一
A x
dx (Leibniz7 notation)
/ ;(x〇) (Newton^s notation)
W h e n }^ou use Leibniz7 notation, }^ou have to r e m e m b e r w h e r e 3^ou7re evaluating the derivative
in the e x a m p l e above, at x = x 〇.
Other, equall}^ valid notations for the derivative of a function / include
df
/ ;, a n d D f
5
Lecture 1
18.01 Fall 2006
Example 2. f(x) = xn where n = 1,2, 3...
W h a t is -f-xn ?
ax
T o find it, plug y = f ( x ) into the definition of the difference quotient.
A y
A x
(x 〇 + A x ) n — Xq
(x + A x ) n — x n
A x
A x
( F r o m here on, w e replace x 〇 with x, so as to have less writing to do.) Since
(x + A x ) n = (x -\- A x ) ( x + Ax)...(x + A x ) n times
W e c a n rewrite this as
x n + n ( A x ) x n_1 + O ( ( A x ) 2)
O ( A x )2 is short h a n d for “all of the terms with ( A x ) 2，（A x ) 3 , a n d so o n u p to ( A x ) n .” （This is part
of w h a t is k n o w n as the binomial theorem; see your text b o o k for details.)
A y
A x
T a k e the limit:
Therefore,
(x + A x ) n — x n
x n + n ( A x ) ( x n _ 1 ) + O ( A x )2 — x n
A x
A x
= n x n_1 + O ( A x )
lim
A x ^ O A x
= nxn~ x
d n
—— x n == nxn~ 1
dx
This result extends to polynomials. For example,
d
dx (x2 十 3 x 10) = 2 x 十 3 0 x 9
Physical Interpretation of Derivatives
Y o u c a n think of the derivative as representing a rate of c h a n g e (speed is o n e e x a m p l e of this).
O n Halloween, M I T students have a tradition of dropp i n g p u m p k i n s f r o m the roof of this building,
w h i c h is a b o u t 4 0 0 feet high.
T h e equation of m o t i o n for objects near the ea r t h s surface (which w e will just accept for n o w )
implies that the height a b o v e the g r o u n d y of the p u m p k i n is:
y = 4 0 0 - 16t2
T h e average speed of the p u m p k i n (difference quotient) =
At
= ^ stance travelled
time elapsed
W h e n the p u m p k i n hits the ground, y = 0,
400 - 16t2 = 0
6
Lecture 1
18.01 Fall 2006
Solve to find t = 5. T h u s it takes 5 seconds for the p u m p k i n to reach the ground.
A v e r a g e s p e e d :
4 0 0 ft
5 sec
80 ft/s
A spectator is prob a b l y m o r e interested in h o w fast the p u m p k i n is going w h e n it slams into the
ground. T o find the instantaneous velocity at t = 5, lefs evaluate y f\
y f = —32t = (—32)(5) = —160 ft/s (about l l O m p h )
y' is negative because the p u m p k i n ^ y-coordinate is decreasing: it is m o v i n g d o w n w a r d .
7
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