Clase: Derivadas-multiplicación y división. Método de la cadena. Derivadas del logaritmo.

Universidad de Guayaquil Facultad de Ingeniería Química Carrera de Ingeniería Química
Nombre:	María Fernanda Díaz
Asignatura:	Matemáticas 1
Paralelo:	1er semestre B
Fecha:	29/01/2016
Docente:	Ing. Manuel Fiallos

Ejercicios en clase:

d(x^n)
d(u^n)=(nx)^(n-1) du
d(u+v)=du+dv
d(u*v)=udv+vdu
d(u*v*z)=uvdz+uzdv+vzdu
d(u/v)=(vdu-udv)/v2

En caso de multiplicación

y=(x+2)(x+5)
y´=(x+2)d(x+5)+(x+5)d(x+2)
y´=(x+2)1+(x+5)1
y´=2x+7

y=(x+2)(x+5)
y=x^2+7x+10
y=d(x^2 )+d(7x)+d(10)
y´=2x+7

y=(x^2+2)(x^2+5)
y´=(x^2+2)(2x)+(x^2+5)(2x)
y´=(2x)((2x^2+7)
y´=4x^2+14x

Y=(3x+2)(x^2+5)
y´=(3x+2)(2x)+(x^2+5)(3)
y´=(6x^2+4x +3x^2+5)
y´=9x^2+4x+5

y=(3/x+2)(2x^3+5)
y´=(3/x+2)(6x^2 )+(2x^3+5)(-3/x^2 )
y´=18x+12x^2-6x-15/x^2
y´=12x+12x^2-15/x^2

y=(3/x+2)(2x^(1/2)+5)
y´=(3/x+2)(x^(-(1/2) ) )+(2x^(1/2)+5(-3/x^2 )
y´=(3/x^(3/2) +2/x^(1/2) )+(-6/x^(3/2) -15/x^2 )
y´=-3/x^(3/2) +2/x^(1/2) -15/x^2

En caso de divisiones

d(u/v)=(v*du-u*dv)/v^2
y=(x+2)/(x+5); x≠-5
y´=((x+5)(1)+(x+2)(1))/(x+5)^2
y´=3/(x+5)^2

y=(x^2+2)/(x+5)
y´=((x+5)(2x)-(x^2+2)(1))/(x+5)^2
y´=(2x^2+10x-x^2+2)/(x+5)^2
y´=(x^2+10x+2)/(x+5)^2

y=(x^2+4)/(x+2)
y=(x+2)(x-2)/(x+2)
y=x-2
y´=1

y=(x^2+4)/(x+2)
y´=((x+2)(2x)-(x^2+4)(1))/(x+2)^2
y´=(2x^2+4x-x^2+4)/(x+2)^2
y´=(x^2+4x+4)/(x+2)^2
y´=(x+2)^2/(x+2)^2
y´=1

y=(x^3–a^3)/(x-a)
y=(x-a)(x^2+ax+a^2 )/(x-a)
y=x^2+ax+a^2
y´=2x+a

y=(3x^2+5)/(x+2)
y´=((x+2)(6x)-(3x^2+5)(1))/(x+2)^2
y´=(6x^2+12x-3x^2+5 )/(x+2)^2
y´=(3x^2+12x+5)/(x+2)^2

Método de la cadena

y=u^n
y´=n*u^(n-1)*du
y=(x+2)^2
y´=2(x+2)^(2-1)*d(x+2)
y´=2x+4

y=(x^2+a)^2
y´=2(x^2+a)d(x^2+a)
y´=4x(x^2+a)
y´=4x^3+4xa

y=((x+2)/(x-2))^3
y´=3((x+2)/(x-2))^2*d((x+2)/(x-2))
d((x+2)/(x-2))=((x-2)(1)-(x+2)(1))/(x-2)^2
d((x+2)/(x-2))=-4/(x-2)^2
y´=(-12(x+2)^2)/(x-2)^4

y=((x+2)(x+3)/(x-2))^4
y=4((x+2)(x+3)/(x-2))^3 d((x+2)(x+3)/(x-2))
d((x+2)(x+3)/(x-2))=((x-2)(2x+15)-x^2-5x-6)/(x-2)^2
d((x+2)(x+3)/(x-2))=x^2-4x-16
y´=(4(x+2)^3 (x+3)^3 (x2-4x-16))/(x+2)^5

y=[((x+2)+(x+3)^2)/(x+2)]^4
y´=4[((x+2)+(x+3)^2)/(x+2)]^3 d[((x+2)+(x+3)^2)/(x+2)]^4
d[((x+2)+(x+3)^2)/(x+2)]^4=1-1/(x+2)^2
y´=4[((x+2)+(x+3)^2)/(x+2)]^3 (1-1/(x+2)^2)
y´=(4(x+1)(x+3) (x(x+7)+11)^3)/(x+2)^5

y=(√(x+2))/(x-2)
y´=((x-2)(d√(x+2)-√(x+2))/(x+2)^2
y´=((x-2)/(2√(x+2))-√(x+2)/(x+2)^2)
y´=((x-2)-2(x+2))/(2√(x+2))
y´=(x-2)/(2√(x+2))-2(x+2)/(2√(x+2))
y´=(x-2)/(2√(x+2)( x-2)^2)-(x+2)/((2√(x+2))(x-2)^2)
y´=1/(2 √((x+2)(x-2)))-(x+2)/(2√((x+2) (x-2)^2))

Derivadas del logaritmo
d(lnx)=1/X d(ln u)=d u/u

y=ln⁡(x+2)^2

y´=d((x+2)^2)/(x+2)^2
y´=2(x+2)d(x+2)/(x+2)^2
y´=2(x+2)/(x+2)^2
y´=2/(x+2)
y=ln⁡ x^3+3/x^2
y´=d(ln⁡ x^3)+d(3/x2)
y´=d(x^3)/x^3-6x^(-3)
y´=(3x^2)/x^3–6x^(-3)
y´=3x^(-1)-6x^(-3)

y=ln√(x^2+3x)

y´=(d(x^2+3x)^(1/2))/(x^2+3x )^(1/2)
y´=(1/2(x^2+3x)^(-1/2) (2x+3))/(x^2+3)^(1/2)
y´=((2x+3)/(2√(x^2+3x)))/(√(x^2+3x))
y´=(2x+3)/2(x^2+3)
y´=(2x+3)/(2 (x^2+3))

y=√(ln⁡(x^2+4x))

y=[ln⁡(x^2+4x)]^(1/2)
y´=1/2 [ln⁡(x^2+4x ]^(-1/2) d(ln⁡(x^2+4x))
y´=(d(ln x{2+4x)))/(2√(ln⁡(x^2+4x))))
y´=((2x+4)/(x^2+4x))/(2√(ln⁡(x^2+4x)))
y´=(x+2)/(ln⁡(x^2+4)(x^2+4x))