Post on 04-Apr-2018
7/30/2019 Basic Derivatives Formulas
1/4
7/30/2019 Basic Derivatives Formulas
2/4
7/30/2019 Basic Derivatives Formulas
3/4
Linear Approximationf(x)
f (x0) + f( x0) (x x0)
Newtons Method xn+1 =xn f(xn )
f (xn ), start with initial value x0
LHopital rule for indeterminate limits ( 00
, or
): limxc f(x)g(x)
= limxc
f (x)g (x)
=L .
For 00,1,
0
limxc[ f(x)]g(x) , let y= [ f(x)]g(x ), then lny =g(x) ln f(x); or y = eg(x )ln f(x)
Application1.Domain, vertical asymptotes, horizontal asymptotes (Test 1)2.Continuity, differentiable at a point (check left /right side)
3.Application of Derivatives a. Slope of tangent b. Rate of change, or velocityc. Increase /Decrease (y ) , Concavity (y ) d. Max/min (y =0 or undefined)
Optimization (Sec 3.7) e) Sketch graph (section 3.6, handouts, follow procedure)
4. Application of Integration a. Area * b. Distance c. Average value(Integration Mean
Value Theorem)Also you write down your own formulas.
Sec 5.5 Integration by Substitution
1. Substitution in Indefinite Integrals Ex. Evaluate 2xex2
dx
Choose a new variable
u = x2;then
du
dx= u du = u dx = 2xdx , and dx =
du
2x, 2xex
2
dx =
2xex2
eu
du
2x = eudu =
eu
+ c = ex2
+ c . (most time, replace dx and do cancellation)
2.
f[g(x)] g (x)dx = f(u)du ,
u = g(x),du = g (x)dx ,
u:inner most function of integrand
Ex1. (3sinx + 4)5 cosxdx Let u =3 sin x + 4, then du =( 3sinx +4 ) dx = 3cosx dx,
dx =du
3cosx,
(3sinx + 4)5cosxdx = (3sinx + 4)5
u5
cosx du3cosx =1
3u
5du =
1
3u
6+ c=..
Ex2.sin x
xdx, let u =
x , du = (x1
2 ) dx =1
2x
1
2dx =1
2 xdx , so dx = 2 xdu, and
7/30/2019 Basic Derivatives Formulas
4/4
sin x
xdx =
sin x
sin u
x 2 xdu = 2 sinudu =
= 2cosu+ c = 2cos x + c
Ex3.(tan
1x)
2
1+ x2 dx , since
d
dx(tan
1x) =
1
1+ x2dx . Letu =
tan1
x, du =1
1+ x2dx, and
(tan1x)
2
1+ x2 dx = (tan
1
x)2
u2
1
1+ x2 dx
du
= u2
du =1
3 u3
+ c =
1
3(tan1
x)3
+ c ( not solve dx!)
3.Substitution in Definite Integrals: x 2x2 +1dx0
2
, u = 2x2 +1
du = 4xdx,dx =du
4x. find the
new limits of integration for u . When
x= 0,
u = 2(0) +1=1;
x = 2,u = 2(4) +1= 9
So
x 2x2+1dx = x u
1
9
0
2
du
4x=
1
4u
1
2du1
9
=1
42
3u
3
2 |19=
1
6(9
3
2 1)=
1
6(3
31) =
26
6=
13
3
4.sin3
cos3+ 2d,
0
/ 3
u = cos3+ 2, du = sin(3)d d=du
sin3, and
= 0,u = cos(0)+ 2 =1+ 2 = 3;= /3,u = cos()+ 2 = 1+ 2 =1
sin3
cos3+ 2d= sin(3
)
u
du
3sin3= 1
31udu
3
1
= [ 13 lnu] |31=
3
1
0 / 3
( 13 ln1+ 1
3ln 3) = 1
3ln 35.(#42)
x
1+ x4dx , if use u =1+ x 4,du = 4x 3dx , dx =
du
4x3
, then
x
1+ x4dx =
x
u
du
4x3=
1
4
1
x2
1
ud u (unsuccessful!) (
1
x2
1
u d
u , or1
x2dx
1
u d
u)
If let u = x2,du = 2xdx , and note that x4 = (x2)2 = u2,
x
1+ x4dx =
x
1+ u2du
2x =
1
2
1
1+ u2du =
1
2tan
1(u) =
1
2tan
1(x
2)+ c
Also fore
t
1+ e2t0
1
,u = e t,du = e tdt,e2t = (e t)2 = u2[0,1][1,e ]
du
1+ u21
e
=
tan1u |
1
e