basic derivatives formulas

7/30/2019 Basic Derivatives Formulas

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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


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

basic derivatives formulas

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