DU DC Y - Y0 m= X - X0 m=
m( X - X 0 ) = Y - Y0 3 8 39 b= 8
m( X - X 0 + Y0 = Y
m=
3 ( X - 3) + 6 = Y 8 3 9 X - +6 =Y 8 8 3 39 X+ =Y 8 8
Y=
3 39 X+ 8 8
m( X - X 0 ) = Y - Y0 m( X - X 0 + Y0 = Y - 4( X - (-5)) + 8 = Y - 4 X - 20 + 8 = Y - 4 X - 12 = Y
x –12 -4X – 12 = Y
DU DC Y - Y0 m= X - X0 m=
-1- 4 -3- 2 m =1 m=
m( X - X 0 ) = Y - Y0 m( X - X 0 + Y0 = Y 1( X - 2) + 4 = Y X -2+4 =Y X +2 =Y
æ3 7ö Pç , ÷ è2 5ø æ-6 9ö P2 ç , ÷ è 5 7ø
æ3 9ö ç ,- ÷ è2 5ø
æ 8 7ö ç - ,- ÷ è 3 2ø æ 10 3 ö ç ,- ÷ è 9 7ø
-
15 7
-
5 8
y
y 6
6 5
5 4
4
3
3
2
2
1
1
x
x -5
-4
-3
-2
-1
1
2
3
4
5
6
-2
-1
1 -1
-2
-3
-3
-4
-4
3
2
1 x -2
-3
-2
y
-3
-4
-1
4
-4
-5
-1
1 -1
-2
-3
-4
2
3
4
5
2
3
4
5
6
CĂĄlculo, La Derivada
f (x) lim xÂŽa
y = f ( x) = 2 x - 1
15
f (x)
lim g ( x) = 17
g ( x) = 5 x - 3
lx ® 4
(5 x - 3) - 17 <
x-4 <
(5 x - 3) - 17 <
5x - 20 <
5( x - 4) < x-4 <
x-4 <
Î 5
Î 5
Î 5
Î ¶
x
g(x)
X
3.90
16.50
4.1
17.5
3.95
16.75
4.05
17.25
3.98
16.90
4.02
17.10
3.99
16.95
4.01
17.05
17
lim g ( x)
17
lim g ( x) x®4
g(x)
x®4
Î= 17 - 16.50 = 0.50
Î= 17.5 - 17 = 0.50
¶ = 4 - 3.90 = 0.10
¶ = 4.1 - 4 = 0.10
Î 0.50 = =5 ¶ 0.10
Î 0.50 = =5 ¶ 0.10
Î= 17 - 16.75 = 0.25
Î= 17.25 - 17 = 0.25
¶ = 4 - 3.95 = 0.05
¶ = 4.05 - 4 = 0.05
Î 0.25 = =5 ¶ 0.05
Î 0.25 = =5 ¶ 0.05
0 ¥ k ¥ - ¥;0 x¥; ; ; ;0 0 ; (+¥) 0 ;1± ¥ 0 ¥ 0
¥
lim f ( x) = lim K = K x®a
x®a
lim 25a = 25a x®2
lim x®a
[f ( x)] = éê ë n
lim x®a
ù f ( x)ú û
f ( x) = (4 x + 5)3
n
3
lim (4 x + 5)
3
x ® -2
é ù 3 (4 x + 5ú = [4(-2) + 5] = (-3)3 = -27 ê ë x®-2 û
lim
lim Kf ( x) = K lim f ( x) x®a
x®a
lim 6 x = 6 lim x = 6(-4) = -24 x ® -4
x ® -4
lim [f ( x) ± g ( x)]= lim f ( x) ± lim g ( x) x®a
lim ( x x®2
x®a
2
+ 5 x - 4) =
x®a
2
lim x + lim 5x - lim 4 = (2) x®2
x®2
lim [f ( x).g ( x)]= lim f ( x).lim g ( x) x®a
x®a
x®a
x®2
2
+ 5(2) - 4 = 10
lim ( x + 2)(3x - 4) = lim ( x + 2).lim (3x - 4) = (1 + 2)(3(1) - 4) = -3 x ®1
lim x®a
x ®1
é f ( x) ù ê ú= ë g ( x) û
x ®1
lim f ( x) lim g ( x) x®a
x®a
lim x®2
é 4x + 3ù ê 2x -1 ú = ë û
lim (4 x + 3) 4(2) + 3 11 = = lim (2 x - 1) 2(2) - 1 3 x®2
x®2
lim
n
f ( x) = n
x®a
lim
3
lim f ( x) x®a
(3 x + 5) = 3
x ®1
lim [f ( x)]
é =ê ë
lim x®a
ù lim x ®a f ( x)ú û
(3(1) + 5) = 3 8 = 2
g ( x)
é ù 3x lim [4 x - 2] = êlim (4 x - 2)ú x®2
3
x ®1
g ( x)
x®a
lim (3x + 5) =
ë
x®2
û
3x lim x ®2
= [4(2) - 2]
3( 2 )
= 6 6 = 46656
lim log
b
x®a
é f ( x) = log b ê ë
ù
lim f ( x)úû x®a
é
ù
lim log ( x + 1) = log êëlim ( x + 1)úû = log (26 + 1) = log (27) = 3 3
3
x ® 26
3
3
x ® 26
lim f ( x) = P
lim f ( x) = lim f ( x) = P x®a -
x®a
x®a
x®a +
x ® a+
-
lim ( x x®2
2
+ 1) =
lim ( x x®2
2
-
+ 1) =
lim ( x x®2
2
+ 1) = 5
+
g (x ) g (x )
f (x )
x
f (x )
x
lim x ® -2
( x 2 - 4) x+2
0 0
lim x ® -2
( x - 2)( x + 2) = x+2
lim ( x - 2) = -4 x ® -2
lim x ®1
x -1 x -1
0 0
lim x ®1
x -1 = x -1
lim x ®1
( x - 1)( x + 1) = x -1
1 x
x ® ¥+
x ® ¥-
x -1
lim ( x - 1)( x ®1
x + 1)
=
lim x ®1
1 x +1
=
1 2
¥- ¬ x ® 0 0 ¬ x ® ¥+
x®¥ 1
lim x = 0 x ®¥ +
x ® ¥+
-
1
lim x = 0 x ®¥ -
1
lim x = -¥ x ®0
-
1
lim x = +¥ x ®0
+
0 ¥ k ¥ - ¥;0 x¥; ; ; ;0 0 ; (+¥) 0 ;1± ¥ 0 ¥ 0
lim (4 x
3
+ 5 x - 8)
x®2
3
lim
3x + 2
5
2x - 4
lim
8 x 3 - 27 2x - 3
x ® -1
x®
3 2
lim h ®3
lim h ® -1
lim t ®1
lim x ®0
h+6 h-6 2h 2 - h - 3 h +1
1- t 5 - t2 - 2 1 1 x+2 2 x
lim x ®0
x+5 - 5 x 4
lim f ( x) = 3 x®d
2
lim g ( x) = - 7 x®d
lim [6 g ( x)] x® d
lim [2 f ( x) + 3g ( x)] x®d
lim x® d
lim x®d
7 f ( x) 2 g ( x) 3 f ( x) + 14 g ( x)
lim Dx ®0
(h + Dh) 2 - h 2 Dh 4x2 -1
lim 2 x + 1 x®-
1 2
lim x ®¥
lim
2 x 3 - x 2 + 3x + 1 3x 3 + 2 x 2 - x - 1
ln k
x-2
x®2
x-2
lim
x2 -1 5 + 4x2
x ®¥
t- 5 2 -5
lim t t® 5
m-h - h+m h
lim h ®0
lim x ® -5
æ 8 x 3 + 125 ö ç ÷ ç x+5 ÷ è ø x-c
lim ac - x(c + a) + x
2
x ®c
2
é1 - n ù
lim êë n + 1 úû
n 2 +3 -2 n 2 -1
n ® -1
lim x®
é
1 3
30 x 2 + 8 x - 6 2(3 x - 1)
1ù
lim êë1 + n úû x ®¥
lim
n
x ®0
=e
ex -1 x
Dx,
dy , y ' , f ' ( x) dx
T Q(x,y)
f(x) = y
f(x0) = y0 θ
m=
P
(x0,y0 )
α x0
y - y 0 f ( x ) - f ( x0 ) = x - x0 x - x0
m = tan gq =
f ( x ) - f ( x0 ) x - x0
lim θ = α Q→p
lim tanθ = tanα Q→p f ' ( x) =
lim DC ®0
f ( x + Dx) - f ( x) DC
f ( x) =
f ' ( x) =
f ' ( x) =
lim DC ®0
f ( x + Dx) - f ( x) Dx
lim
1 1 x + Dx x = Dx
DC ®0
- Dx
= lim xDx( x + Dx) DC ®0
lim DC ®0
=
x - ( x + Dx) x( x + Dx) Dx
=
-1
lim x( x + Dx) DC ®0
1 (2, ) 2 1 1 = - (2) + b 2 4
1 y = - x +1 4
lim DC ®0
1 x
x - x - Dx x( x + Dx) Dx
DC = 0 = -
1 x2
y 4
3
2
1 x 1
mN =
-1 mT
y
2
3
4
5
( mT .m N = -1 ) Normal tangente
y0
X X0
mT =
lim X ®0
mT =
f ( x + Dx) - f ( x) = Dx
3 x - 3 x - 3Dx
lim xDx( x + Dx) X ®0
mT (3) = -
lim
= lim X ®0
X ®0
3 3 x + Dx x = Dx
3Dx = xDx( x + Dx)
1 3
m T ( 3) = -
lim X ®0
3 x - 3( x + Dx) xDx( x + Dx) 3
lim x( x + Dx) X ®0
-
1 3
m N (3) =
1 3
1 3
y = mx + b 1 y =- x+b 3
1 y=- x+2 3
y = -mx + b 1 y = x+b 3
1 y= x 3 y
Recta tangente
YT =
1 X+2 3
Recta normal
YN =
1 X 3
1
F(x) = 3
6
x
3 X
f ( x) = 5 x - 3
f ( x) = 2 x 2 - 3x + 5 g (t ) =
2 t
1 2
y= x
z=
x2 + 1 x +1
y=
1 x2
f (q ) =
1
q f ( z ) = ( z + 1) 3
f (t ) = (t + 1)(t - 1)
M = ( z 2 - 4 z + 3)
DK = 0 d K =0 dx
Lectura: La derivada de una constante es igual a cero
f ( x) = 10
y =3 y' = 0
y=
f ' ( x) = 0
1 3
y=z dy =0 dx
dy =0 dx
DX = 1 d x =1 dx
y=x y '= 1
Lectura: La derivada de X con respecto a X es igual a uno
y=z dy =1 dz
f (t ) = t f ' (t ) = 1
dy ยบ y ' ยบ D ยบ f ' ( x) dx Lectura La derivada de una constante por una variable o funciรณn es igual a la constante multiplicada por la derivada de la variable o funciรณn.
t 10 1 f ' (t ) = Dt 10 1 f ' (t ) = (1) 10 1 f ' (t ) = 10 f (t ) =
y = 3x y ' = 3Dx y ' = 3(1) y' = 3
y = 3x y ' = 3Dx + xD3 y ' = 3(1) + x(0) y' = 3
D
u vDu - uDv = v v2
4 t tD 4 - 4 Dt y' = t2 t (0) - 4(1) y' = t2 4 y' = - 2 t y=
Du n = n(u ) n -1 Du
y = x4 y ' = 4( x) 4-1 Dx y ' = 4( x) 3 (1) y' = 4 x3
v Du v = u v [ Du + Ln(u ) Dv] u
y = x Lnx Lnx Dx + Lná xñ Dx] x Lnx y ' = x Lnx [ (1) + Lnx (1)] x Lnx y ' = x Lnx [ + Lnx ] x y ' = x Lnx -1 Lnx (1 + x) y ' = x Lnx [
D(u ± v) = Du ± Dv
y = 3x + x 3 - 6 y ' = D(3 x) + Dx 3 - D6 y ' = 3Dx + 3( x) 3-1 Dx - 0 y ' = 3(1) + 3 x 2 (1) y ' = 3(1 + x 2 )
f ( x) =
f ' ( x) =
x2 + 2x x3 - 5 x 3 - 5 D( x 2 + 2 x) - ( x 2 + 2 x) D x 3 - 5 ( x3 - 5 )2
1
x 3 - 5 ยบ ( x 3 - 5) 2 1
f ' ( x) =
x 3 - 5[ Dx 2 + D 2 x] - ( x 2 + 2 x) D( x 3 - 5) 2 ( x3 - 5 )2 1
f ' ( x) = f ' ( x) =
1 x 3 - 5 (2 x + 2) - ( x 2 + 2 x) ( x 3 - 5) 2 D( x 3 - 5) 2 ( x 3 - 5)
2(2 x + 2)( x 3 - 5) - ( x 2 + 2 x)3 x 2 3
2( x 3 - 5) 2 f ' ( x) =
4( x + 1)( x 3 - 5) - 3 x 3 ( x + 2) 3
2( x 3 - 5) 2
1 y = 5x 4 - x3 + 8x 2 - 9 3 1 x -2 y = 9t -3 - t - 4 + 5 4
z = x 3 + 85 x
f (t ) = 3 8t -
3 x3
f ( x) = x 4 ( x 3 + 5)
g ( x) = x -3 ( x 2 - 8) 4 y=
8x 2 3x - 4
z=
3t -2 + 5t 3 6x3 + 2x
g ( x) =
x2 - 5 6 - x3
x = t2
x = t2 d (x) dt
x' = 2(t ) 2-1.Dt x' = 2t (1) x' = 2t
x' = 2(1) Þ x' = 2
x = 4t 2 + 3t y = t 3 + 2t
y=
4t 3 - 5t 2t - 1
y = x3 + 2x 2 + 3
y = x3 + 2x 2 + 3 y ' = 3x 2 + 4 x y' ' = 6 x + 4 y' ' ' = 6
y ' , f ' ( x),
dy d , [ f ( x)], Dx ( y ) dx dx
y ' ' , f ' ' ( x),
d2y d2 , 2 [ f ( x)], D 2 x ( y ) 2 dx dx
y ' ' ' , f ' ' ' ( x),
d3y d3 , [ f ( x)], D 3 x ( y ) dx 3 dx 3
y ( n ) , f ( n ) ( x),
d (n) y d (n) , [ f ( x)], D ( n ) x ( y ) dx ( n ) dx ( n )
y ( 5)
f ( x) = x 2 f (t ) = 2 -
2 t
f ( z) = z
V (t ) =
20t t +1
dy dy du = dx du dx
dy =1 du
du = 3u 2 dx
dy dy du = dx du dx
dy = 1(3u 2 ) dx dy = 3( x + 1) 2 dx
du dx 1
y = (5 x 2 - 3) 2
y=
y=
5 3
3
x +2
1 (x 2
y=-
- 3) 5
4 ( x + 5) 3
dy du
dy dy du = dx du dx
y=
6 x2
dy , y ' , f ' ( x), Dx ( y ) dx dy dx dy dx dy dx dy dx
dy d = (3 y 2 + 5 y - x 3 ) = (3) dx dx dy dy 2 dx 3(2 y + 5 - 3 x =0 dx dx dx dy dy 6 y + 5 - 3x 2 = 0 dx dx
dy dx 6y
dy dy + 5 = 3x 2 dx dx
dy dx dy [6 y + 5] = 3 x 2 dx
dy 3x 2 = dx 6 y + 5
dy dx
Y
-1
X -2
x2 + 9 y2 = 9
( 6, -
1 3
)
3( x 2 + y 2 ) 2 = 100 xy dy dx
x 2 + y 2 = 36 9 x 2 + 16 y 2 = 144 4 y2 - x2 = 4
4 x - y 3 + 12 y = 0
1
f ( x) = x 2
y = x2
h( x) = x -2
f ( x) = 4 x
f ( x) = K x f ' ( x) = K x .Ln K
d x dx
DK u = K u .Ln K .u '
d u du K = K u .Ln K dx dx
Lectura
d u du e = e u .Ln e dx dx Lectura
De u = e u .u '.
v Du v = u v [ u '+ ln u v' ] u Lectura
y = x Lnx Lnx Dx + Lná xñ Dx] x Lnx y ' = x Lnx [ (1) + Lnx (1)] x Lnx y ' = x Lnx [ + Lnx ] x y ' = x Lnx -1 Lnx (1 + x) y ' = x Lnx [
Ln x = b Ln x
eb = x
Ln x. y = Ln x + Ln y Ln
x = Ln x - Ln y y
Ln x y = yLn x
Ln e x = x e
Ln x
=x
y = Log b x by = x y = Log e x º y = Ln x
Log b u =
Ln u Ln b
Log 315 =
Ln 15 Ln 3
=
2.70805 » 2.46498 1.09861
Cรกlculo, La Derivada
Log 5 53 Log 41 Log10 x 3 y 1 x3 x Log 5 ( 2 ) y
Log 3
e Lnt = 5
Ln z = 0
e3 x+ 2 = 6 Ln x 3 = 27
5(3t ) = 10
f ( x) = Log b u d 1 dU ( Log bU ) = . dx U .Ln b dx D( Log bU ) =
U' ULn U
f ( x) = Ln u d 1 dU ( Ln U ) = . dx U dx U' D( Ln U ) = U Lectura
dy dx x2 + 2 ) x -1 x+2 y = Ln( Ln( )) x-2 y = xx
Ln( x 3 + 3) y= x3 3 y = Ln(e Log 2 ( x - 2 x ) )
y = Lob3 (
3
y = 6 x .8 LnX y = Ln x
x
y = ( x 3 + 2)
LnX
x2
y = x 3 + Lnx 3 (10 Log 4e ) 1
y = Lnx 2
lim sen(u + Du ) - senu d ( senu ) = Du ® 0 dx Du =
= =
lim
senu cos Du + cos u.senDu - senu Du ® 0 Du lim
senu (cos Du - 1) - cos usenDu Du ® 0 Du lim Du ® 0
senu (
Lim cos Du senDu )+ cos u D u ® 0 Du Du
= senu (0) + cos u (1)
= cos u
Lim senx =1 x®0 x Lim 1 - cos x =0 x®0 x
y = sen( x 2 ) dy d = cos( x 2 ). ( x 2 ) dx dx dy = cos( x 2 ).2 x dx dy = 2 x cos( x 2 ) dx
3
dy dx
y = sen( x 2 ) 3
dy dx
y = sec(e 2 ) y = Ln cos( x y =5
y=
2
1 + 3) 2
tan g ( t 3 + Lnt )
sen( x 2 + 3) - tan gx cos( x 3 )
dy dt
dy dx
dy dx
d u' (arc.senu ) = dx 1- u2 d u' = (arc. cos u ) = dx 1- u2 d u' (arc.Tangu ) = dx 1+ u2 d u' (arc.ctngu ) = dx 1+ u2 d u' (arc. sec u ) = dx u u2 -1
d u' (arc. csc u ) = dx u u2 -1
e3 x
D(arc.sen(e 3 x ) =
D (e 3 x ) 1- e
=
6x
3(e 3 x ) 1 - e6 x
e3 x
x x
2
h(t ) = arc.sen(e ( x 3+ ln x ) )
y = x arc .cos( x
3 2
)
z = arc. tan g ( x x ) y = arc.sen(
x3 + 5x - 3 1 3x 3
)
U V
f ( z) =
z +1 z
4 t2 y = t +1
M=
y = 6x3 - 2x - 6
y = x 3e x
h( x ) =
( x + 3) 2 x -1
f (q ) = sen(q + 3). cos 3 (4q )
f (t ) = (t 3 - 4t )5t
r -1
y=
r +1
2
f (u ) =
e (u +1) u 3 + 5e cos u
y = arc.Tang (u 3 + 5u ). cos(u 2 + 2) y = ( x 3 + 3) e
f (t ) = t +
y = ex
1 t
6 x +3
x3 + 5x - 2 6x + 1
y=3 y=
y=
arc.sen(t + 1) arc. cos(t + 1)
sen(t + 2) (t + 1) (t
y = x2 - x y= x
2
-1)
3
y = (4 x 3 + 5) 2
y = senx
y = arc. sec(u 3 )
y=
t 3 + 5t - 2 t -1
1
y = (6 x 3 + 5) 2 f ( x) = e
x3 +3
5x3 y 2 + 2 y 3 = 6x sen( xy ) + cos x 2 - tan g ( y 3 ) = 6 y 5x 2 + 2 y 3 xy
= 8 xy
Ln( x 2t ) + 5 xt = 8 xt
y=
1 3
2
t +1 1 3 Q( x) = cos( x + 5) 3
A = 2pr 2 + 2prs
V = pr 2s
s=
V pr 2
dA =0 dr
dV cm 3 = 50 dt s
10 pr 2 10 10 2 A = 2pr + 2pr[ 2 ] A = 2p (r 2 + ) pr pr dA 10 = 2p (2r - 2 ) dr pr s=
dr = 10cm dt
V =
4 pr 3 3 dV dV dr dV = . = 4pr 2 dt dr dt dt dr 1 dV = dt 4pr 2 dt dr 1 = (50 ) dt 4p (10 ) 2 1 cm 8p s
dr dt dr 1 = dt 8p
x
h x
A = x 2 + 4 xh x 2 + 4 xh
300 - x 2 4x 300 - x 2 V = x2 ( ) 4x dV x3 =0 V = 75 x dx 4 x = Âą10 x = 100 h=
h = 20t 2 + 10
0 = 75 -
3x 2 4
dr dt
r h
5000 pies
Por relación de Pitágoras tenemos que r 2 = (5000)2 + h2 calculando la velocidad de la nave por medio de la derivada de la ecuación dada, la dh cual nos da como resultado = 40t, como nuestra incógnita está en dt dr cuando t = 20 segundos y vemos que r y h se relacionan según dt la función r 2 = (5000) 2 + h 2 ,ahora bien derivando implícitamente respecto a t, tenemos: dr dh Despejando dr 2r = 2h dt dt dt dr h dh dh dr h Reemplazando = = 40t , no queda = (40t ) cuando dt r dt dt dt r t = 20 segundos en la función h = 20t 2 + 10 , da como resultado h = 20(20) 2 + 10 por lo tanto h = 8010 pies, luego r = (5000)2 + (8010)2 donde r = 9442.46 pies. Finalmente dr pies =678.64 dt s
dr 8010 = (40)(20) , entonces dt 9442.46
22 x 8 tan gl = x
22 ) x 8 l = arc. tan g ( ) x 22 8 q = ar. tan g ( ) - arc. tan g ( ) x x tan gd =
14
δ ϴ
8
λ
d = arc. tan g (
x
22 8 - 2 2 dq x x = 2 2 dx æ 22 ö æ8ö 1+ ç ÷ 1+ ç ÷ è x ø è xø -
22 8 2 2 = 2 x + 2x x + 484 x + 64 x2 x2 -
8 22 = 2 x + 64 x + 484 2
7 2 x = 420 4 x = 240
dq =0 dx
x 2 + 484 =
11 2 x + 64 4
(
)
dy dt
dx =4 dt 1 V = pr 2 h 3
C
dC : dx dR : dx dG : dx
C( x) = a + bx + ex 2 + kx3
C ( x) = 300 + 0.06 X + 2 x10 -4 X 2
C ( x) = 300 + 0.06 X + 2 x10 -4 X 2 300 + 0.06 X + 2 x10 -4 X 2 x C ' ( x) = 0.06 + 0.0004 x
C ( x) =
Unidades
Costo
Costo medio
Costo marginal
X
C(x)
c(x)
Câ&#x20AC;&#x2122;(x)
400
356
0.89
0.22
1200
660
0.55
0.54
4000
3740
0.935
1.66
C ( x) = 300 + 0.06(400) + 2 x10 -4 (400) 2 C ( x) = 356 C ( x) 356 C ( x) = = x 400 C ( x) = 0.89 C ' ( x) = 0.06 + 4 x10 -4 (400) C ' ( x) = 0.22 C ' ( x) = c( x)
C ' ( x) = c( x)
0.06 + 0.0004 x =
C ( x) despejando x
x(0.06 + 0.0004 x) = 300 + 0.06 x + 0.0002 x 2 0.06 x + 0.0004 x 2 = 300 + 0.06 x + 0.0002 x 2 0.0002 x 2 = 300 300 0.0002 x = 1225unidades x=
300 + 0.06(1225) + 2 x10 -4 (1225) 2 1225 C ( x) = 0.55 C ( x) =
C ( x) =
C ' ( x) = c( x) x2 - 20 x 30 x C ' ( x) = - 20 15 C ( x) =
C ' ( x) = c( x) x - 20 = 0 15 x = 300
x2 - 20 x 30
R( x) = XP( x)
3x + D 2 = 21000
D = 21000 - 3 x 0 ยฃ x < 7000 21000
D' =
-3 2 21000 - 3 x
D (pesos)
Grรกfica de la funciรณn demanda 100
50
5000
7000
x unidades
R( x) = x 21000 - 3x
R' ( x) =
x=
42000 - 9 x 2 21000 - 3 x
42000 @ 4667 9 0 £ x < 4667 4667 < x < 7000
P ( 4667 ) = 21000 - 3( 4667 ) @ $83.66
U ( p ) = 0.6 p 2 + 20
p (t ) = 5.1 + 0.2t 2
dU dt 1 1 dU 1 = 0.6 p 2 + 20 2 [1.2 p ] = 0.6 p 0.6 p 2 + 20 2 dt 2
(
)
(
)
dp = 0.4t dt
1 dU dU dp dU 0.24 pt = . Þ = 0.6 p 0.6 p 2 + 20 2 .(0.4t ) = dt dp dt dt 0.6 p 2 + 20
(
2
p (4) = 5.1 + 0.2(4 ) = 8.3 dU 0.24 pt 0.24(8.3)(4) = = 2 dt 0.6 p + 20 0.6(8.3) 2 + 20
dU 7.968 = = 1.017 dt 61.334
)
1 x 2
1 x2
2000
350 x
-1
p (t ) = 30 - 6(t + 2 )
A(t ) = 0.4 p 2 + p + 70
x 10 x -1
x -1
S = 40t - t 2
1 2 gt 2
S ( p) =
6.45 p2
p (t ) = 0.03t 2 + 0.2t + 8
C ( x) = 5 x 2 + x + 60
.
4 5
4 5
2 5 2 10
lim x ®0
x+5 - 5 x
5 2 1 5 1
lim x - 8 - x x ®8
2
72 - 64
1 8 1 16 1 4
h( x) =
lim f ( x) = a x ®¥
4 + x2 16 - x 2
lim f ( x) = a x ® -¥
lim f (x )= f (c ) x ®c
f (c )
lim f (x )¹ f (c ) x ®c
p(x )= 3500 +
1200 x +1
(h + Dh) - h Dh 速0 Dh f (h + Dh) - h lim Dh 速0 Dh f (h + Dh) - f (h) lim Dh 速0 Dh lim
lim h 速0
f (h + Dh) - f (h) Dh
3m 2 s 3m 2 s 2m 3 s 2m 3 s
1 P( x) = 900 X 3
P (t ) = -t 3 + 9t 2 + 12t
op. hip. ady. cosq = hip. op. tan gq = Opuesto ady.
ady. op. hip. secq = ady. hip. tan gq = op.
senq =
Hipotenusa
q
c tan gq =
Adyacente
RELACIÓN POLAR
RELACIÓN RADIAN
(GRADOS)
RELACIÓN RECTANGULAR
0
0
30
p 6
45
p 4
60
p 3
1 2 ( , ) 2 2
90
p 2
(0,1)
1 2 (- , ) 2 2
2p 3
(1,0)
( (
3 1 , ) 2 2 2 2 , ) 2 2
y
r b ?
a x
x
tan gq = r= a2 + b2
1 csc x 1 cos x = sec x senx tan gx = cos x senx =
cos x senx 1 sec x = cos x 1 csc x = senx c tan gx =
sen 2 x + cos 2 x = 1 sec 2 x - tan g 2 x = 1 csc 2 x - c tan g 2 x = 1
1 - cos 2u 2 1 + cos 2u cos 2 u = 2 1 - cos 2u tan g 2u = 1 + cos 2u sen 2u =
b r a cosq = r senq =
(a,b)
c tan gq = r a r cscq = b
secq =
b a
a b
sen(2u ) = 2 senu cos u cos(2u ) = cos 2 u - sen 2u = 2 cos 2 u - 1 = 1 - 2 sen 2u tan g (2u ) =
2 tan gu 1 - tan g 2u
sen(u Âą v) = senu. cos v Âą cos u.senv cos(u Âą v) = cos u.senu m senu.senv tan g (u Âą v) =
tan gu Âą tan gv 1 m tan u. tan gv
u+v u -v ). cos( ) 2 2 u+v u -v senu - senv = 2 cos( ).sen( ) 2 2 u+v u -v cos u + cos v = 2 cos( ). cos( ) 2 2 u+v u -v cos u - cos v = -2 sen( ).sen( ) 2 2 senu + senv = 2 sen(
1 senu.senv = [cos(u - v) - cos(u + v)] 2 1 cos u cos v = [cos(u - v) + cos(u + v)] 2 1 senu cos v = [ sen(u + v) - sen(u - v)] 2 1 cos u.senv = [ sen(u + v) - sen(u - v)] 2