ࡱ> { p4y S T U $%VW*HIJ (!)!*!"""M$N$%%%&h'i'^(_(U)V)W)X)Y)Z)v,w,x,y,z,{,|,//0000000k4l4m4n4 bjbjVV 1<<b"  II9)9)9)D}+}+}+.1}+͈3HRUHUHI\h\h\hLNNNNNN$ӊu`r9)\he\h\h\hrIIUHI|||\hI UH9)IL|\hL|| Pn%(I`,3}+kv80͈w6ՍyՍՍ9)4\h\h|\h\h\h\h\hrr|\h\h\h͈\h\h\h\hՍ\h\h\h\h\h\h\h\h\h : DERIVACIN DE FUNCIONES III.1. INCREMENTOS Definicin.- El incremento (x de una variable x es el aumento o disminucin que experimenta desde un valor x = x0 a otro x = x1 de su campo de variacin. As pues, (x = x1 - x0. Si se da un incremento (x a la variable x ( es decir, si x pasa de x = x0 a x = x0 + (x ), la funcin y = f(x) se ver incrementada en (y = f ( x0 + (x ) f(x0 ) a partir del valor y = f(x0 ). Ejemplo 13: Si se tiene que y = f(x) = x2 Si x0 = 10, entonces fija a y = 100 Suponiendo x1 = 12, entonces fija a y = 144 Resulta que (x = 2, determina (y = 44 Suponiendo x1 = 9, entonces fija a y = 81 Resulta que (x = -1, determina (y = -19 El cociente  EMBED Equation.3  =  EMBED Equation.3  recibe el nombre de cociente medio de incrementos de la funcin en el intervalo comprendido entre x = x0 hasta x = x0 + (x III.2. DERIVADAS Notacin usual en las derivadas.- La derivada de una funcin de una variable es el lmite de la razn del incremento de la funcin al incremento de la variable independiente cuando este ( el incremento (x ) tiende a cero. Cuando el lmite existe, se dice que la funcin es derivable o tiene derivada. Matemticamente:  EMBED Equation.3   EMBED Equation.3 =  EMBED Equation.3  EMBED Equation.3  la derivada ser el lmite del segundo miembro cuando (x ( 0 y se representa  EMBED Equation.3  que se lee: la derivada de y [ o de f(x) ] con respecto a x  EMBED Equation.3  =  EMBED Equation.3   EMBED Equation.3  si u es una funcin de t  EMBED Equation.3 =  EMBED Equation.3   EMBED Equation.3  las expresiones  EMBED Equation.3  y  EMBED Equation.3  deben considerarse como un todo y no como una fraccin. Si y = f(x), la derivada se expresa de diferentes formas, algunas de ellas son:  EMBED Equation.3  =  EMBED Equation.3  f(x) =  EMBED Equation.3  y = f ( (x) = y ( = Dx y = Dx f(x) Regla general de derivacin.- Para encontrar la derivada de una funcin conforme a la definicin anterior, se recomiendan los pasos siguientes: Se sustituye en la funcin x por x + (x y se calcula y + (y. Se resta el valor inicial de la funcin del valor obtenido y + (y para encontrar el incremento (y. Se divide el incremento de la funcin ((y) sobre el incremento de la variable ((x). Se calcula el lmite de este cociente cuando el incremento de la variable ((x) tiende a cero. El lmite encontrado es la derivada buscada. Ejemplo 14: Encontrar la derivada de y = 3x2 + 5 I.- y + (y = 3 ( x + (x )2 + 5 y + (y = 3x2 + 6x(x + 3 ((x)2 + 5 II.- y + (y = 3x2 + 6x(x + 3 ((x)2 + 5 ( y = ( (3x 2 + 5 (. (y = 6x(x + 3 ((x)2 III.-  EMBED Equation.3  =  EMBED Equation.3  = 6x + 3(x IV.-  EMBED Equation.3  =  EMBED Equation.3   EMBED Equation.3  =  EMBED Equation.3  (6x + 3 (x ) = 6x EJERCICIOS VI Encuentra la derivada de cada funcin que se proporciona. y = 2 3x f(x) = mx + b y = x4 ( =  EMBED Equation.3  s = at2 + bt +c y =  EMBED Equation.3  f(x) = cx3 y = 3x x3 y = x2 + 2x s = ( a + bt )2 y =  EMBED Equation.3  y = x2 + 2  EMBED Equation.3   EMBED Equation.3   EMBED Equation.3  III.3. INTERPRETACIN GEOMETRICA DE LA DERIVADA Geomtricamente, la derivada de una funcin f(x) es la pendiente de la recta tangente a la grfica de y = f(x) en el punto [ x0, f(x0) ]. En la figura siguiente: considerando que: y = f(x) I.- y + (y = f ( x + (x ) NQ II.- y + (y = f ( x + (x ) NQ ( y = ( f(x) MP = NR . (y = f( x + (x) - f(x) RQ III.-  EMBED Equation.3  =  EMBED Equation.3  =  EMBED Equation.3  = tg ( RPQ = tg ( = pendiente de la secante PQ. La razn del incremento (y al incremento (x es la pendiente de la secante determinada por P ( x, y ) y Q ( x + (x, y + (y ). IV.- Si se considera x como fijo, entonces P es punto fijo en la grfica. Si (x vara tendiendo a cero, el punto Q se mueve en la curva y se acerca a P como lmite. La recta PQ gira sobre P y se sobrepone a PT. ( = inclinacin de la secante PQ t = inclinacin de la tangente PT  EMBED Equation.3 ( = t y ( =  EMBED Equation.3 tg ( = tg t = pendiente de la tangente en P. El valor de la derivada en cualquier punto de una curva es igual a la pendiente de la tangente a la curva en aquel punto. Ejemplo 15: Hallar la pendiente m , el ngulo de inclinacin ( y la ecuacin de la recta tangente a la curva y = 3x2 + 5 en los puntos (0, 5) y (  EMBED Equation.3 , y ). Por lo anterior:  EMBED Equation.3  = m  EMBED Equation.3 = EMBED Equation.3  ( 3x2 + 5 ) = 6x ; m = 6x; resultado del ejemplo 12 para x = 0, m = 6 ( 0 ) = 0 m = 0 como m = tg (, tg ( = 0 ( = arc tg (0) = 00 ( = 00 para (0, 5) y m= 0; la Ec. es y 5 = 0 ( x 0 ); y 5 = 0 para x =  EMBED Equation.3 , m = 6 ( EMBED Equation.3 ) = 2 m = 2 ( = arc tg (2) = 63.430 ( = 63.430 para ( EMBED Equation.3 ,  EMBED Equation.3 ) y m= 2; la Ec. es y  EMBED Equation.3  = 2 ( x  EMBED Equation.3 ); 6x 3y + 14 = 0 EJERCICIOS VII Encuentra la pendiente m , el ngulo de inclinacin ( y la ecuacin de la recta tangente en cada caso: y = x2 2 ( 1, -1 ) b) y = 2x -  EMBED Equation.3 x2 ( 3, y ) c) y =  EMBED Equation.3  x = 2 d) y = 3 + 3x x3 x = -1 e) y = x3 x2 ( -1, -2 ) III.4. DERIVACIN DE FUNCIONES ALGEBRAICAS Debido a que la aplicacin de la regla general de derivacin ( por incrementos ) sera muy laboriosa, en su lugar, existen frmulas de derivacin. Para derivar funciones algebraicas, las frmulas correspondientes son las siguientes:  EMBED Equation.3  = 0 siendo c una constante  EMBED Equation.3  = 1  EMBED Equation.3  =  EMBED Equation.3  +  EMBED Equation.3  -  EMBED Equation.3   EMBED Equation.3  = c  EMBED Equation.3  siendo c una constante  EMBED Equation.3  = u  EMBED Equation.3  + v  EMBED Equation.3   EMBED Equation.3  = u v EMBED Equation.3  + uw  EMBED Equation.3  + vw  EMBED Equation.3  donde u, v, w son funciones derivables de x  EMBED Equation.3  = n v n 1  EMBED Equation.3   EMBED Equation.3  = n x n 1  EMBED Equation.3  =  EMBED Equation.3   EMBED Equation.3  =  EMBED Equation.3  siendo c ( 0 Ejemplo 16: Derivar la funcin de cada inciso: y = 3x2 + 5x3 + x4  EMBED Equation.3  = EMBED Equation.3  (3x2 + 5x3 + x4 ) =  EMBED Equation.3  (3x2 ) +  EMBED Equation.3  ( 5x3 ) +  EMBED Equation.3  (x4 ) aplicando 3) u( = 3 EMBED Equation.3  (x2 ); v( = 5 EMBED Equation.3  (x3 ); w( =  EMBED Equation.3  (x4 ); aplicando 4) u( = 3 (2x2-1 ); v( = 5 (3x3-1 ); w( = 4x4-1); aplicando 8) y( = 6x + 15x2 + 4x3 s = ( t2 3 )4  EMBED Equation.3  = EMBED Equation.3  [( t2 3 )4 ] ; u = t2 3; n = 4 s( = 4 ( t2 - 3 )4-1  EMBED Equation.3 ( t2 3 ); aplicando 7) s( = 4 ( t2 - 3 )3 [ EMBED Equation.3 ( t2 )  EMBED Equation.3 ( 3 ) ]; aplicando 3) s( = 4 ( t2 - 3 )3 [ 2t 0 ]; aplicando 8) y 1) s( = 8t ( t2 - 3 )3 f(x) =  EMBED Equation.3 ; f (x) =  EMBED Equation.3  f ((x) =  EMBED Equation.3   EMBED Equation.3  f ((x) =  EMBED Equation.3   EMBED Equation.3  EMBED Equation.3   EMBED Equation.3  aplicando 7) f ((x) =  EMBED Equation.3   EMBED Equation.3   EMBED Equation.3  aplicando 9) f ((x) =  EMBED Equation.3   EMBED Equation.3   EMBED Equation.3  aplicando 1) y 2) f ((x) =  EMBED Equation.3   EMBED Equation.3   EMBED Equation.3  = EMBED Equation.3  f ((x) =  EMBED Equation.3 ; f ((x) =  EMBED Equation.3  EJERCICIOS VIII Encontrar la derivada de la funcin en cada inciso: y = x5 + 5x4 - 10x2 + 6 f(x) = 3x1/2 - x3/2 + 2x-1/2 y =  EMBED Equation.3  +  EMBED Equation.3  y =  EMBED Equation.3  f(t) =  EMBED Equation.3  +  EMBED Equation.3  y = ( 1 5x )6 f(x) = ( 3x x3 + 1 )4 f(z) =  EMBED Equation.3  y = 2x2  EMBED Equation.3  ( =  EMBED Equation.3  y =  EMBED Equation.3  y =  EMBED Equation.3  z =  EMBED Equation.3  s =  EMBED Equation.3  s = t  EMBED Equation.3  f(t) = ( 2 3t2 )3 y = x ( a + bx )1/2 y =  EMBED Equation.3  s =  EMBED Equation.3  r =  EMBED Equation.3  y =  EMBED Equation.3  f(x) =  EMBED Equation.3  f(t) = at5 5bt2 s = 2t4/3-3t2/3 III.5. DERIVADAS SUCESIVAS La derivada de una funcin tambin es una funcin, ello nos permite intentar derivarla repetidas veces. As, f( (x), f(((x), f((((x), f4(x), ..................., fn (x) denotarn la primera, segunda, tercer, cuarta, ........, n-sima derivada de la funcin f. Ejemplo 17: Si f(x) = x5 2x3 ; tendremos: f ( (x) = 5x4 6x2; f (((x) = 20x3 12x; f ((((x) = 60x2 - 12, f 4(x) = 120x; f 5(x) = 120 y de ah en adelante todas las derivadas sern igual a cero. Algunas otras formas de expresar derivadas sucesivas son: f (((x) = f 2(x) =  EMBED Equation.3  = y ((; f ((( = f 3(x) =  EMBED Equation.3  = y ((( y as sucesivamente. EJERCICIOS IX Calcular las derivadas indicadas en cada problema: y = 3x4 2x2 + x 5 y((( = ? y =  EMBED Equation.3  y(IV) = ? f(x) =  EMBED Equation.3  f (((x) = ? y =  EMBED Equation.3  y(( = ? y =  EMBED Equation.3  y(V) = ? y =  EMBED Equation.3  y(( = ? y = x2 4x + 8 y((( = ? III.6. DERIVACIN DE FUNCIONES IMPLICITAS Cuando se da una relacin entre x y y por medio de una ecuacin no resuelta para y [ f(x, y) =0], entonces y se llama funcin implcita de x ( o tambin x funcin implcita de y ). Por ejemplo: x2 4y = 0 A veces, es posible resolver la ecuacin que define una funcin implcita con respecto a una de las variables, obteniendo as una funcin explcita. As, puede definirse y como funcin explcita de x: y =  EMBED Equation.3 x2. Sin embargo, puede ocurrir que la resolucin indicada sea imposible o complicada; cuando sucede tal caso, para calcular la derivada de esta clase de funciones se aplican los siguientes pasos: Derivar trmino a trmino con respecto a x y donde aparece y derivarla como funcin de x. Agrupar trminos con  EMBED Equation.3 en el primer miembro. Despejar  EMBED Equation.3  Ejemplo 18: Calcular la derivada de: 15x = 15y + 5y3 + 3y5  EMBED Equation.3  (15x) =  EMBED Equation.3  (15y) +  EMBED Equation.3  ( 5y3 ) +  EMBED Equation.3  (3y5 ) 15 = 15  EMBED Equation.3  + 15y2  EMBED Equation.3  + 15y4  EMBED Equation.3   EMBED Equation.3  [ 1 + y2 + y4 ] = 1;  EMBED Equation.3  =  EMBED Equation.3  Ejemplo 19 Derivar: x3 3axy + y3 = 0  EMBED Equation.3  (x3) -  EMBED Equation.3  (3axy) +  EMBED Equation.3  ( y3 ) =  EMBED Equation.3  (0) 3x2 3ay 3axy( + 3y2y( = 0 y( (3y2 3ax ) = 3ay 3x2 ; y( =  EMBED Equation.3  EJERCICIOS X Encuentra la derivada de cada funcin implcita que se proporciona: x2 + y2 = 4 y3 + xy 10 = 0 x2 + 3y2 4 = 0 x2 - 2xy + y2 = 6 y3 + 3x2y + x2 2xy = 3 x2 + y2 4x + 6y 24 = 0 x =  EMBED Equation.3  x2 + xy + 2y2 = 28 x2/3 + y2/3 = a2/3 x + xy + y = 2 x2 xy + y2 = 3 x2 xy2 + x2 + y2 = 0 Hallar la pendiente de las curvas indicadas en los puntos sealados: x3 2xy + y3 = 5; P( 1, 1 ) y3 + (y x )2 = 7 + x y P( 1, 2 ) x2 + y2 = 4 P( 2, 0 ) x2 y + xy2 = 12 P( 3, 1 ) x2 - y2 = 3 P( 2, 1 ) 2y3 + 4xy + x2 = 7; P( 1, 1 ) x3 - y3 = 5xy - 3 P( 2, 1 ) 2y3 + 4xy + x2 = 7 P( 1, 1 ) III.7. DERIVACIN DE FUNCIONES TRASCENDENTES. El siguiente grupo de frmulas de derivacin se aplican para derivar funciones trascendentes (llamadas as para distinguirlas de las algebraicas vistas anteriormente). Las funciones trascendentes se dividen en : Funcin logartmica y exponencial. Funcin trigonomtrica o circular. Funcin trigonomtrica inversa o circular inversa. (1) Funciones logartmicas y exponenciales.- Las frmulas para derivar estas funciones son:  EMBED Equation.3  =  EMBED Equation.3  =  EMBED Equation.3   EMBED Equation.3   EMBED Equation.3  =  EMBED Equation.3   EMBED Equation.3   EMBED Equation.3  = a v ln a  EMBED Equation.3   EMBED Equation.3  = e v  EMBED Equation.3   EMBED Equation.3  = v u v-1  EMBED Equation.3  + ln u (uv)  EMBED Equation.3  Ejemplo 20: Calcular la derivada de la funcin proporcionada en cada inciso: y = ln ( 4x 5 ); v = 4x + 5 ln v = ln ( 4x 5 ) y( =  EMBED Equation.3  EMBED Equation.3  (4x 5 ) aplicando 11) y( =  EMBED Equation.3  (4 0 ) ; y( =  EMBED Equation.3  f(x) = e5x f((x) = e5x  EMBED Equation.3  (5x ) aplicando 14) f((x) = e5x (5 ) ; f((x) = 5 e5x y = x2 ex y( = ( x2 )  EMBED Equation.3  ( ex ) + (ex )  EMBED Equation.3  ( x2 ) y( = ( x2 ) [ ex  EMBED Equation.3  (x ) ] + (ex ) [ 2x ] y( = x2 ex (1 ) + 2x ex ; y( = ex ( x2 + 2x ) y =  EMBED Equation.3  y( =  EMBED Equation.3   EMBED Equation.3  ( ex ) y( =  EMBED Equation.3  ( ex ); y( =  EMBED Equation.3  y = 22x y( = 22xln 2  EMBED Equation.3  ( 2x ) aplicando 13) y( = 22xln 2 ( 2 ) y( = 2 ( 22x ) ln 2 EJERCICIOS XI Calcula la derivada de cada funcin indicada: y = ln ( x2 + a ) y = x2 ln x2 y = b  EMBED Equation.3  f(x) =  EMBED Equation.3  y = ln ( 3x2 + 5 ) f(x) = ln ( ln x ) s = e-t y =  EMBED Equation.3  y = ex ln x y = ln ( x2 ex ) f(x) =  EMBED Equation.3  f(x) =  EMBED Equation.3  y =  EMBED Equation.3  y = ( ln x2 )3 y = ln  EMBED Equation.3  y =  EMBED Equation.3  f( x) =  EMBED Equation.3  s = t et f(()=b2(  EMBED Equation.3  (2) Funciones trigonomtricas o circulares.- En la derivacin de funciones trigonomtricas circulares, las frmulas que se aplican para resolver estos problemas son:  EMBED Equation.3  = cos v  EMBED Equation.3   EMBED Equation.3  = - sen v  EMBED Equation.3   EMBED Equation.3  = sec2 v  EMBED Equation.3   EMBED Equation.3  = - csc2 v  EMBED Equation.3   EMBED Equation.3  = sec v tg v  EMBED Equation.3   EMBED Equation.3  = - csc v ctg v  EMBED Equation.3  NOTA: El argumento de estas funciones trigonomtricas se expresa en radianes. Ejemplo 21: Calcular la derivada de: y = 4 tg 5x y( =  EMBED Equation.3  ( 4 tg 5x ) ; y( = 4  EMBED Equation.3  ( tg 5x ) y( = 4 [ sec2 5x  EMBED Equation.3  ( 5x )] aplicando 18) y( = 4 sec2 5x ( 5 ); y( = 20 sec2 5x sen y = cos 2x funcin implcita.  EMBED Equation.3  ( sen y ) =  EMBED Equation.3  ( cos 2x ) cos y  EMBED Equation.3  = - sen 2x  EMBED Equation.3  ( 2x ) aplicando 16) y 17) cos y y( = -2 sen 2x; y( =  EMBED Equation.3  y =  EMBED Equation.3  ctg 8x y( =  EMBED Equation.3  [ - csc2 8x  EMBED Equation.3  ( 8x ) ] aplicando 19) y( =  EMBED Equation.3  ( 8 ) ( - csc2 8x ) ; y( = - 2 csc2 8x EJERCICIOS XII Encuentra la derivada de cada funcin que se proporciona: y = sen 3x2 f(x) = 3x2 4 cos x s = tg 3t y = 4 ctg  EMBED Equation.3  f(t) =  EMBED Equation.3  + 7 cos t y = sen ax2 f(() = 3 cos 2( y = tg 3 4x f(() = cos3 ( y = sen 2x cos 2x y = a csc bx s = tg ( - ( f(() =  EMBED Equation.3  g(x) = sec x tg x y =  EMBED Equation.3  sen x + cos y = 0 f(() =  EMBED Equation.3  s =  EMBED Equation.3  y = x sen x y = EMBED Equation.3  Encuentra la derivada de las funciones implcitas que se proporcionan: 2 sen x + 4 cos y = 4 sen 2y cos 2x = 0 ctg y = x y sen x + cos y = ey cos 2 y = tg 3 x (3) Funciones trigonomtricas inversas o circulares inversas.- La funcin inversa de x = sen y es y = arc sen x (y = sen-1 x). Para las dems funciones trigonomtricas, su funcin inversa se encuentra de la misma manera. Las frmulas de derivacin de estas funciones son:  EMBED Equation.3  =  EMBED Equation.3   EMBED Equation.3  = - EMBED Equation.3   EMBED Equation.3  =  EMBED Equation.3   EMBED Equation.3  = - EMBED Equation.3   EMBED Equation.3  =  EMBED Equation.3   EMBED Equation.3 = -  EMBED Equation.3  Ejemplo 22: Encontrar la derivada en cada caso: y = arc sen ( 2x 3 ) y( =  EMBED Equation.3  [ arc sen ( 2x - 3 ) ] y( =  EMBED Equation.3  EMBED Equation.3  ( 2x - 3 ) aplicando 22) y( =  EMBED Equation.3  ( 2 ) y( =  EMBED Equation.3  ; y( =  EMBED Equation.3  y = arc csc 2x y( =  EMBED Equation.3  [ arc csc 2x ] y( = -  EMBED Equation.3  EMBED Equation.3  (2x ) aplicando 27) y( = -  EMBED Equation.3 ; y( = -  EMBED Equation.3  y2 sen x + y = arc tg x  EMBED Equation.3  ( y2 sen x ) +  EMBED Equation.3  ( y ) =  EMBED Equation.3  (arc tg x ) y2 ( cos x ) + sen x ( 2y ) y( + y( =  EMBED Equation.3  EMBED Equation.3  ( x ) y( ( 2y sen x + 1) =  EMBED Equation.3  ; y( =  EMBED Equation.3  EJERCICIOS XIII Calcula la derivada de las funciones de cada inciso. y = arc tg 2x2 y = arc cos  EMBED Equation.3  y = arc cot  EMBED Equation.3  y = arc sec  EMBED Equation.3  f(x) = arc sen ( 3x 4x3 ) y = x arc sen 2x y = x2 arc cos x y = arc csc mx y = arc sen ( a x2 + b x + c ) f ( x ) = x arc tg x y = arc sen  EMBED Equation.3  encontrar y ( para cada valor de x y = x arc sen x x =  EMBED Equation.3  y = x arc cos x x = -  EMBED Equation.3  y =  EMBED Equation.3  x = 1 y = x2 arc csc  EMBED Equation.3  x = 2 V ANLISIS DE FUNCIONES. IV.1. LA DERIVADA EN LAS FUNCIONES CRECIENTES Y DECRECIENTES Funcin creciente.- Una funcin es creciente si y [ f(x) ] aumenta cuando x aumenta. Su derivada es positiva. Funcin decreciente.- Una funcin es decreciente si y disminuye cuando x aumente. Su derivada es negativa. Grficamente:  Al variar un punto a lo largo de la curva de izquierda a derecha, la curva sube ; es decir, a medida que x aumenta, la funcin ( y ) aumenta. Los incrementos (y y (x son del mismo signo.  Al moverse el punto de izquierda a derecha sobre la curva, esta baja ; a medida que x aumenta, y ( la funcin ) disminuye. En este caso los incrementos (y y (x son de signos opuestos. Una funcin puede ser unas veces creciente y otras decreciente. Con la funcin:  y = 2x3 9x2 + 12x 3; cuya grfica es: asumiendo que un punto se mueve de izquierda a derecha a lo largo de la curva se pueden hacer las siguientes consideraciones: hasta llegar a A la curva sube, desde A hasta B la curva baja, y de B en delante, continua subiendo. Es decir: Desde x = - ( hasta x = 1 la funcin es creciente Desde x = 1 hasta x = 2 la funcin es decreciente Desde x = 2 hasta x = + ( la funcin es creciente Generalizando lo anterior: Una funcin es creciente cuando su derivada es positiva ( pendiente positiva, ngulo agudo ) y decreciente cuando su derivada es negativa ( pendiente negativa, ngulo obtuso ). Para la funcin del caso: y = 2x3 9x2 + 12x 3 = 6 ( x 1 ) ( x 2 ) y = 6 x2 18 x + 12 = 6 ( x2 3 x + 2 ) = 6 ( x 1 ) ( x 2 ) Cuando x ( 1; f ((x) es positiva; f(x) es creciente Cuando 1 ( x ( 2; f ((x) es negativa; f(x) es decreciente Cuando x ( 2; f ((x) es positiva; f(x) es creciente Resultados congruentes con la representacin grfica. Ejemplo 23: Para f(x) = 6 x2 , investigar el intervalo donde la funcin es creciente y donde es decreciente. f ((x) = - 2x Para valores negativos de x ( x ( 0 ), f((x) ser positiva ( es decir, creciente ); cuando a x se le asignen valores positivos ( x ( 0 ), f((x) ser negativa ( es decir, decreciente ). Resumiendo: Cuando x ( 0; f((x) es positiva; f(x) es creciente Cuando x ( 0; f((x) es negativa; f(x) es decreciente  La grfica correspondiente es: EJERCICIOS XIV Encuentra los intervalos donde cada funcin proporcionada sea creciente o decreciente: y = x2 - 9 y = x2 y = 2x - x2 y = 2x2 - x4 y = x3 6x2 + 9x y = 5x y = ( x 3 )3 y = 3x2 x3 IV.2. MXIMOS Y MINIMOS. Definicin.- El valor de una funcin es un mximo si es mayor que cualquiera de los valores que le anteceden o le siguen inmediatamente. Un valor de una funcin es un mnimo si es menor que cualquiera de los valores inmediatamente anteriores y posteriores al considerado. En la grfica siguiente y = 2x3 9x2 + 12x 3 la funcin tiene un valor mximo MA ( y = 2 ) en x = 1 y un valor mnimo NB ( y = 1 ) en x = 2. A los anteriores conceptos se les llama mnimos y mximos locales, ya que en la grfica se observa que a la derecha de B hay valores de y mayores que MA y a la izquierda de A tambin la funcin tiene valores menores que el mnimo NB. En el punto A de la grfica, y( = 0 ( m = 0, tangente del ngulo de cero grados es cero ); en el punto C a la izquierda de A, el ngulo de la tangente es agudo ( pendiente positiva ) y en el punto D a la derecha de A el ngulo de la tangente es obtuso ( pendiente negativa ). En el punto A ( mximo ), la funcin cambia de creciente a decreciente; la derivada de la funcin cambia de positiva ( + ) a negativa ( -). En el punto B, y( = 0; en punto D la funcin es decreciente y en E es creciente. En B ( mnimo ), la derivada de la funcin cambia de negativa ( - ) a positiva ( + ). Lo anterior se puede resumir de la siguiente forma: f(x) es un mximo si f ((x) = 0 y f ((x) cambia de + a f(x) es un mnimo si f ((x) = 0 y f ((x) cambia de a + Los valores que satisfacen la ecuacin f((x) = 0 se llaman valores crticos y determinan puntos de cambio donde la tangente es paralela al eje x. En la funcin del caso, los valores crticos son x = 1 y x = 2. Para determinar el cambio de signo en la derivada, se evala la funcin para valores un poco menores y un poco mayores. Si y( = 6x2 18x + 12: Para x = 1; valor poco menor = 0.5 valor poco mayor = 1.5 y( = 6 ( 0.5 )2 18 ( 0.5 ) + 12 = 1.5 - 9 + 12 ( + y( = 6 ( 1.5 )2 18 ( 1.5 ) + 12 = 37.5 - 27 + 12 ( El cambio es de + a Para x = 2; valor poco menor = 1.5 valor poco mayor = 2.5 y( = 6 ( 1.5 )2 18 ( 1.5 ) + 12 = 37.5 - 27 + 12 ( y( = 6 ( 2.5 )2 18 ( 2.5 ) + 12 = 37.5 - 45 + 12 ( + El cambio es de a + Para x = 1; Mx. = y = 2 Para x = 2; Mn. = y = 1 Primer mtodo para calcular mximos y mnimos.- En la aplicacin de este mtodo, se recomiendan los pasos siguientes: Se encuentra la primera derivada. Se iguala la primera derivada a cero y se calculan las races reales de la ecuacin resultante. Estas races sern los valores crticos de la variable. Se consideran los valores crticos uno por uno, y se calculan los signos de la primera derivada, en primer lugar para un valor un poco menor ( mayor que cualquier valor crtico menor al que se analiza ) que el valor crtico y despus para un valor un poco mayor. Si el signo de la derivada es primeramente + y despus , la funcin tiene un mximo para este valor crtico de la variable; en caso contrario, tiene un mnimo. Si no hay cambio de signo, no hay mximo ni mnimo para ese valor crtico. Ejemplo 24: Calcular mximos y mnimos de f (x) = x3 6x2 + 9x I)  EMBED Equation.3  f ( (x) = 3x - 12x + 9 II) 3x - 12x + 9 = 0 x - 4x + 3 = 0 (x-3) (x-1)= 0 Las races son: x = 1; x = 3. (se pueden calcular con la frmula general) III) Para el valor crtico x = 1: Valor poco menor = 0; Valor poco mayor = 2. f ( (0) = 3(0) - 12(0) + 9 = 9 ( + f ( (2) = 3(2) - 12(2) + 9 =-3 ( el cambio de signo de la derivada es de + a ; por lo que se concluye que hay un mximo. Mx. = f (1) = (1) - 6 (1) + 9(1) = 1-6+9 = 4 Para el valor crtico x = 3 Valor poco menor = 2; Valor poco mayor = 4 f ( (2) = ( f ( (4) = 3(4)-12(4)+9=9 ( + el cambio de signo de la derivada es de a + ; por lo que se concluye que hay un mnimo. Mn. = f (3) =(3)-6(3)+9(3) = 54 - 54=0  Resumiendo: Para x = 1; Mx. = 4 Para x = 3; Mn. = 0 La grfica correspondiente es: Mximos o mnimos para f (x) continua y f (x) se vuelve infinita.- En la grfica siguiente:  En los puntos B y G (mximos) y en E (mnimo), f(x) es continua pero f (x) se vuelve infinita, ya que la tangente en dichos puntos es paralela al eje y. Por lo anterior se deben de incluir como valores crticos los valores de x para los que f (x) se vuelve infinita; matemticamente, los valores de x que satisfacen la ecuacin  EMBED Equation.3  Obsrvese en la figura que tambin en el punto A, f((x) se vuelve infinita, pero en la funcin no hay mximo ni mnimo. EJERCICIOS XV Calcula los mximos y mnimos en la funcin que se proporciona en cada inciso: f(x) = 10 + 12x 3x2 - 2x3 y = 2x3- 9x2 + 12x 3 f (x) = x3 + 2x2 - 15x 20 y = 2x3 + 3x2 + 12x 4 f (x) =  EMBED Equation.3  y = x4+ 2x3 - 3x2 - 4x + 4 f(x) = (x-1)2 (x+1)3 y = x2 + 2x 3 f ( x ) = 2x3 + 3x2 - 12x f ( x ) = 3x4 - 4x3 - 12x2 y = ( 2 + x )2 ( 1 x )2 f (x) = x3 +  EMBED Equation.3  Concavidad de una curva.- Si el punto P (x, y) describe una curva, la pendiente de la tangente en P varia. Cuando la tangente a P queda debajo de la curva (fig. a), el arco es cncavo hacia arriba, si la tangente queda arriba de la curva (fig. b), el arco es cncavo hacia abajo.  El criterio para determinar el sentido de la concavidad de una curva en un punto es: La grfica de y = f(x) es cncava hacia arriba si la segunda derivada de y con respecto a x es positiva; es cncava hacia abajo (convexo) si esta derivada es negativa. Segundo mtodo para determinar mximos y mnimos.- Las condiciones para mximos y mnimos de f(x) correspondientes a valores crticos de la variable son: f(x) es un mx. si f ( (x) = 0 y f (( (x) es negativa. f(x) es un mn. si f ( (x) = 0 y f (( (x) es positiva. Y se recomiendan los pasos siguientes en el clculo de mximos y mnimos con este mtodo: Calcular la primera derivada de la funcin. Igualar a cero la primera derivada y resolver la ecuacin; las races reales son los valores crticos de la variable. Determinar la segunda derivada. Sustituir en la segunda derivada, en lugar de la variable, cada uno de los valores crticos obtenidos. Si el resultado es negativo, la funcin tiene un mximo para ese valor crtico considerado; si el resultado es positivo, la funcin tiene un mnimo. Si f (( (x) = 0; se aplica el primer mtodo, puesto que puede existir un mximo o mnimo para el valor crtico respectivo. Ejemplo 25: Con el segundo mtodo, determinar mximos y mnimos de y = x4 + 2x3 - 3x2 - 4x + 4 f ( (x) = 4x3 + 6x2 - 6x 4 4x3 + 6x2 -6x 4=0 2 (x+2) (2x+1) (x-1) = 0; valores crticos: x = -2; x = -  EMBED Equation.3 ; x = 1. f (( (x) = 12x2+12x-6 para x = -2: f (( (x) = 12 (-2)2 + 12(-2) 6 =48 24 6 ( + Por ser f (( (-2) positiva, hay un mnimo Mn. = f (-2) = (-2)4 + 2(-2)3 - 3(-2)2 -4 (-2) +4 =0 Para x = -1/2: f (( (x) = 12(-1/2)2)+12(-1/2)-6 = 3 6 6 ( Por ser f (( (-1/2) negativa, hay un mximo Mx. = f(-1/2) = (-1/2)4+2(-1/2)-3(-1/2)-4(-1/2) + 4 = EMBED Equation.3  Para x = 1: f (( (x) = 12(1)2 + 12(1) 6 = 12 +12-6 ( + por ser f (( (x) positiva, hay un mnimo Mn. = f(1) = (1)4+2(1)3 - 3(1)2 - 4(1) +4 =1+2-3-4+4 = 0  Resumiendo: Para x = -2; Mn. = 0 Para x = - 1/2; Mx. = 81/16 Para x = 1; Mn. = 0 La grfica correspondiente es: EJERCICIOS XVI Calcula los mximos y mnimos de las funciones de cada inciso: y = 3 + 2x - x2 f (x) = x3 - 6x2 + 9x 8 f(x) = (x2 - 4)2 y = x4 - 4x2 + 4 y = 3x5 - 20x3 IV.3. PUNTOS DE INFLEXIN Definicin.- Un punto de inflexin en una curva es el punto en el cual la curva pasa de cncava a convexa o viceversa. Es el punto que separa arcos que tienen sentido de concavidad opuestos. En la figura, B es un punto de inflexin.  Las condiciones para que exista un punto de inflexin son: f (( (x) = 0 o no este definida f (( (x) cambie de signo Determinacin de puntos de inflexin.-Para encontrar las coordenadas de los puntos de inflexin de una curva, se recomienda aplicar los siguientes pasos: Encontrar f (((x) (segunda derivada) Igualar a cero f(( (x), resolver la ecuacin resultante para considerar las races reales encontradas. Se evala f (( (x), primero para valores de x un poco menores y despus un poco mayores que cada una de las races obtenidas en el segundo paso. Si f (( (x) cambia de signo, entonces se tiene un punto de inflexin. Se evala la funcin original para determinar la ordenada del punto de inflexin en el valor de x considerado. Hay que recordar que: Si f (((x) es positiva, la curva es cncava hacia arriba. Si f (((x) es negativa, la curva es cncava hacia abajo. Ejemplo 26: Determinar intervalos de concavidad y P.I. de la funcin: f(x) = x 3 - 6x2 +9x 8 f ( (x) = 3x2 - 12x +9; f (((x) =6x 12 6x 12= 0 6x = 12; x = 2 valor poco menor = 1; valor poco mayor = 3 f (((1) = 6(1) 12 = -6 f (( (3)= 6(3) 12 = 6 Por haber cambiado de signo hay P.I. f(2) = (2)3-6(2)2+9(2) 8 =8 24+18 8 = -6 La grfica correspondiente es:  P.I. (2,-6) Para x < 2; cncavo hacia abajo Para x > 2; cncavo hacia arriba EJERCICIO XVII Encuentra los puntos de inflexin y la direccin de la concavidad en las funciones de cada caso: f(x) = x2 + 2x 3 f(x) = 3 + 2x x2 y = x3 + 2x2 4x - 8 f(x) = x3 + 2x2 15x - 20 y = ( 2 x )3 y = x3 - 3x2 + 3 f(x)=(x24)2 IV.4. APLICACION DE MXIMOS Y MINIMOS Introduccin.- En los temas anteriores, se estudi como conocer la forma aproximada de la grfica de una funcin mediante la derivada. Estos mtodos se usan para resolver cierto tipo de problemas de geometra, fsica, ingeniera, biologa, etc. donde es necesario minimizar o maximizar alguna cantidad. En todos estos problemas, la dificultad estriba en que hay que representar el problema real mediante un modelo matemtico estableciendo una relacin funcional y = f(x). Una caracterstica de estos problemas es que se refieren a procesos de variacin: kilmetros por hora ( Km/hr ), precio por tonelada ( p/ton. ), kilmetros por litro ( Km/l ), etc. Todas son razones o cocientes que expresan la variacin de una cantidad con respecto a otra; se le llama razn de cambio. La aplicacin de que hablamos, se detalla mediante los siguientes ejemplos: De un cartn rectangular de 4 dm. por 5 dm. se desea construir una caja abierta cuyas dimensiones le den su mxima capacidad, calcular esas dimensiones. Con el auxilio de las siguientes figuras  El volumen de la caja es: ( l ) ( a ) ( h ); l = 5 2x; a = 4 2x; h = x el volumen depende de la longitud x ( relacin funcional ) V (x ) = ( 5 2x ) ( 4 2x ) ( x ) dm3 ; V (x ) = 4x3 18x2 + 20x Restricciones: Por ser longitud, x ser siempre positiva x no debe ser mayor de dos ( la menor dimensin del cartn es cuatro ); 0 < x < 2 La relacin funcional establece que el volumen de la caja depende del valor de x. Algunos pares de valores son: X:0 EMBED Equation.3  EMBED Equation.3  EMBED Equation.3 11 EMBED Equation.3 1 EMBED Equation.3 1 EMBED Equation.3 2V (x):03 EMBED Equation.3 66 EMBED Equation.3 64 EMBED Equation.3 31 EMBED Equation.3 0  La grfica de esta funcin es: Observando la grfica, el problema se resuelve encontrando las coordenadas del punto mas alto de la grfica. En este punto, la tangente a la curva es horizontal ( pendiente cero ). La derivada de V(x) igualada a cero y resolviendo la ecuacin resultante para x nos da la abscisa. Es un problema de mximos y mnimos. V( (x ) = 12x2 - 36x + 20; 12x2 - 36x + 20 = 0; aplicando la frmula general para ecuaciones cuadrticas: x1 = 0.736; x2 = 2.263 x2 ( 2; no cumple con una restriccin, x1 es la solucin y debe ser un mximo. V(((x) = 24x - 36 Para x = 0.736 V(((0.736) = 24 ( 0.736 ) - 36( Por ser V((( x ) negativa, hay un Mx. = 6.5672 Las dimensiones de la caja sern: l = 5 2x = 5 2 ( 0.736 ) = 3.5275 dm.; a = 4 2x = 4 2 ( 0.736 ) = 2.5275 dm. h = x = 0.736 dm. Determinar las dimensiones del pastizal rectangular de rea mxima que puede ser circundado por una cerca de 1 000 metros de longitud. La figura correspondiente es: El rea del rectngulo es: ( l ) ( a ) l = 500 x; a = x; el rea depende de la longitud x ( relacin funcional ) A (x ) = ( 500 x ) ( x ) m2 ; A (x ) = 500x x2 Restricciones: 0 < x < 500; Algunos valores de x y del rea se dan en la tabla siguiente: x:01050100150200250300350400A (x): [ 1 000]04.922.54052.56062.56052.540 La grfica de esta funcin es:  Nuevamente se aprecia que se trata de un problema de mximos y mnimos ( maximizar ) por lo que se calculan estos: A( (x ) = 500 2x ; 500 2x = 0; x = 250 ( debe ser un mximo ) A( ( 240 ) = 500 2 ( 240 ) ( + A( ( 260 ) = 500 2 ( 260 ) ( Con el primer mtodo, pasa de + a ; hay un Mx. = 62 500 m2 Las dimensiones del rectngulo sern: l = 500 x = 500 250 = 250 m. a = x = 250 m. Se desea proyectar una lata para aceite de forma cilndrica que deba contener un litro ( 1 000 cm3 ) de aceite. Determinar las dimensiones de la lata de manera que la cantidad de metal que se gaste en su fabricacin sea mnima. De las figuras siguientes: Se tiene que volumen del cilindro = Area de la base por la altura = Ab h Ab = ( r2 ; V = ( r2 h como V = 1 000; entonces ( r2 h = 1 000; de donde h =  EMBED Equation.3  El menor gasto de material resultar de las menores reas: lateral, inferior y superior que se utilice en la fabricacin: rea del crculo = ( r2; 2 ( r2 = reas superior e inferior el rea lateral es de forma rectangular con l = 2 ( r, a = h; Atotal = 2 ( r2 + 2 ( r h = 2 ( r2 + 2 ( r [ EMBED Equation.3 ] ; A( r ) = 2 ( r2 + [ EMBED Equation.3 ] ser la relacin funcional donde el rea depende del radio de la lata cilndrica. Una tabulacin del rea que depende del radio es: r:012345678A (r):(2 0061 025723600557560593652 cuya grfica es:  De la que se concluye que se trata de un problema de mximos y mnimos ( minimizar ). La solucin es el punto mas bajo de la curva. A(( r ) = 4 ( r - [ EMBED Equation.3 ] ; 4 ( r - [ EMBED Equation.3 ] = 0 4 ( r3 - 2 000 = 0; r3 =  EMBED Equation.3  =  EMBED Equation.3  r =  EMBED Equation.3  = 5.4192; que debe ser un mnimo A((( r ) = 4 ( +  EMBED Equation.3  A((( 5.4192 ) = 4 ( +  EMBED Equation.3  ( + Por ser + ; hay un Mn. = 2 (( 5.4192 )2 + [ EMBED Equation.3 ] Mn. = 553.57 cm2 Las dimensiones deseadas sern: h =  EMBED Equation.3  = 10.838 cm.; r = 5.4192 cm. Dentro de un terreno sin barda, una persona construir un corral rectangular. Si cuenta con 38 metros lineales de material para cercar y si aprovechar una barda del vecino como se muestra en la figura, determina la dimensin h del lado del corral que quedar en la barda del vecino y la longitud b que proporcionarn la mayor rea posible.  longitud a cercar = 2b + h 2b + h = 38; b =  EMBED Equation.3  m rea del rectngulo es: ( b )( h ) el rea depende de h : ( relacin funcional ) A (h ) =  EMBED Equation.3  m2 ; A ( h ) = 19 h - ( h2 ; Restricciones: 0 < h < 19 Algunos valores de h y del rea se dan en la tabla siguiente: h:048121618192022242630A (h):068120156176180180.5180176168156120  La grfica de esta funcin es: Nuevamente se aprecia que se trata de un problema de mximos y mnimos ( maximizar ) por lo que se calculan estos: A( (h ) = 19 h ; 19 h = 0; h = 19 ( debe ser un mximo ) A( ( 18 ) = 19 ( 18 ) ( + A( ( 20 ) = 19 ( 20 ) ( Con el primer mtodo, pasa de + a ; hay un Mx. = 180.50 m2 Las dimensiones del rectngulo sern: h = 19 m; b =  EMBED Equation.3  =  EMBED Equation.3  = 9.5 m EJERCICIO XVIII Resuelve, aplicando mximos y mnimos, los problemas que en cada inciso se plantean. Se desea conocer las dimensiones del rectngulo inscrito en un crculo de cinco centmetros de radio y que sea de mayor rea. Se desea construir un depsito rectangular de base cuadrada abierto por arriba. Debe tener ciento veinticinco metros cbicos de capacidad, si el costo de las caras laterales es de dos mil pesos por metro cuadrado y el fondo es de cuatro mil pesos por metro cuadrado. Determinar las dimensiones del depsito que den el menor costo de fabricacin. Hallar dos nmeros cuya suma sea ciento veinte y de forma que el producto p de uno de ellos por el cuadrado del otro sea mximo. Se desea construir una caja rectangular de base cuadrada abierta por arriba. Calcular las dimensiones que debe tener si se quiere que su volumen sea mximo y se construye con mil doscientos centmetros cuadrados de material Se quiere construir un recipiente cilndrico y de sesenta y cuatro centmetros cbicos de volumen. Encontrar las dimensiones que debe tener para que la cantidad de material usado en su construccin sea mnima. Un mayorista vende zapatos para correr a veinte pesos el par si le piden menos de cincuenta pares. Si le piden mas de cincuenta ( hasta seiscientos ), el precio por par se reduce en .02 pesos multiplicado por el volumen del pedido. Cul es el pedido que produce el mayor ingreso para el mayorista? Un arquitecto desea disear cierto tipo de ventana de tal manera que la parte inferior sea rectangular y la superior sea un tringulo equiltero ( ver figura ). Si cada ventana tiene un rea de tres metros cuadrados, Cules deben ser sus dimensiones para que el permetro sea el menor posible?  Una fbrica de margarina vende su producto en barras que tienen forma de prisma de base cuadrada cuyo volumen es de ciento ocho centmetros cbicos. Determina las dimensiones de la barra que minimizan la cantidad de papel de la envoltura ( las dimensiones con que se gastara menos papel ). Para que un paquete pueda enviarse por correo es necesario que la suma de su longitud con el permetro de su base no exceda de ciento ocho pulgadas. Encuentre las dimensiones de la caja con la base cuadrada de mayor volumen que se pueda enviar por correo.     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