EJERCICIOS Y DEMOSTRACIONES DE LAS DERIVADAS

EJERCICIOS Y DEMOSTRACIONES DE DERIVADAS

Demuestre la identidad cos2x=cos^2 x-sen^2derivando termino a termino la igualdad sen2x=2senxcosx.

f(x)senx=〖cos〗^2 x-〖sen〗^2x

sen2x=2cosx((x+x)/2)2sen((x-x)/2)

sen2x=2cosx 2x/2 2sen2

sen2x=2〖cos〗^2 x2sen^2

Demuestre que:

a) (arcsenx)’+(arccosx)’=0

1/√(1-x^2 )+-1/√(1-x^2 )

((1-1)/√(1-x^2 ))=0

0/√(1-x^2 )=0

0=0

b) (arctanx)’+arccotx)?=0

1/(1+y^2 )+(-1/(1+x^2 ))=0

(1+x^2-1+y^2)/((1+x^2 )(1+x^2))=0

(x^2+y^2)/((1+y^2 )(1+x^2))=0

x^2+y^2=(1+y^2 )(1+x^2)

0=1+x^2 y^2

x=y

y=x

31. Encuentre las derivadas de las siguientes funciones trigonométricas inversas.

a) y=〖(arc cosx)〗^2

y´=2 ⁡(arc cosx) Derivando

y´=2 (arc cos⁡〖x)〗 (1/√(1-x^2 )) Derivada trascendental del arc cos⁡x

y´=(2 (arc cos⁡〖x)〗)/√(1-x^2 ) LCI (R,*)

b) y=arc sen x/a+arc cos √x/b

y´=1/√(1-x^2/a^2 )*(a-x)/a^2 +1/√(1+x^2/b^2 ) (√x/b^2 +1/(2b√x)) derivada trascendental del arc cos⁡x y del arc sen⁡x

y´=(a-x)/((a^2)√(1-x^2/a^2 ))+1/√(1+x^2/b^2 ) ((2x+b)/(2b^2 √x)) regla de la cadena

y´=(a-x)/((a^2)√(1-x^2/a^2 ))+(2x+b)/((2b^2 √x) √(1+x^2/b^2 )) LCI (R,*,+)

c) y=〖(arc tanx)〗^2

lny=ln〖(arctan)〗^2 Propiedad de ln

lny=2*ln(arc tanx)

1/y*dy/dx=2 1/((arc tanx) )*1/(1+x^2 ) Dx (arco tanx) =(-1)/(1+u^2 )

1/y*dy/dx= 2/((arc tanx)(1+x^2))

dy/dx= 2/((arc tanx)(1+x^2))*〖(arctan)〗^2

dy/dx= 2(arc tanx)/((1+x^2))

d) y=arc cos mx/√(x+m)

y^|=-1/√(1+〖(mx/√(x+m))〗^2 ) Dx(arc cosx)=(-1)/√(1-u^2 )

y'= - 1/√(1+〖mx/(x+m)〗^2 )

y^'= -1/√(((x+m)+〖mx〗^2)/(x+m))

e) w=(arc cos u)/u

w^'=(((-1)/√(1-u^2 ))- arc cos u )/u^2

w^'=((-u)/√(1-u^2 )- arc cos u )/u^2

w^'=((-u- arc cos u √(1-u^2 ))/√(1-u^2 ) )/u^2

w^'=(-u- arc cos u √(1-u^2 ) )/(u^2 √(1-u^2 ))* √(1-u^2 )/√(1-u^2 )

w^'=(-u√(1-u^2 )- [〖(√(1-u^2 )〗^2)arc cos u])/(u^2 (√(〖1-u^2〗^2)))

w^'=(-u√(1-u^2 )-[1-u^2 (arc cos u)])/(u^2-u^4 )

w^'=(-u√(1-u^2 )-arc cos u+u^2 (arc cos u))/(u^2-u^4 )

f) y= √(a^2-x^2 )+ a arc sen ax

y^'= (-2x)/(2 √(a^2-x^2 ))+ a(a/√(1-(ax)^2 ))

y^'= -x/√(a^2-x^2 )+ a^2/√(1-a^2 x^2 )

y^'= -(x√(1-a^2 x^2 )+a^2 √(a^2-x^2 ))/((√(a^2-x^2)))*((√(a^2-x^2))(√(1-a^2 x^2 )))/((√(a^2-x^2))(√(1-a^2 x^2 ))) y^'=((x√(1-a^2 x^2 ))(√(a^2-x^2)) (√(1-a^2 x^2 ))+(a^2 √(a^2-x^2 ))(√(a^2-x^2))(√(1-a^2 x^2 )))/((√(a^2-x^2))(√(1-a^2 x^2 )))

y^'= ((x-a^2 x^3 )(√(a^2-x^2 ))+(a^4-a^2 x^2 )(√(1-a^2 x^2 )))/((a^2-x^2 )(1-a^2 x^2))

g) y=arc cos⁡〖x^2 〗 Dx (arc cosx)=(-1)/√(1-u^2 )

y'=((-1)(2x))/√(1-x^4 )

y'=(-2x)/√(1-x^4 )

h) y=√(arc cosx^2 )

y= 〖(arc cos⁡〖x^2)〗〗^(1/2) Dx (arc cosx)=(-1)/√(1-u^2 )

y^'=[1/2 (arc cosx)^((-1)/2) ][(-2x)/(1-x^4 )]

y^'= -1/((1-x^4)√arccosx)

i) u=arctan (v+p)/(1-pv)

u^'=1/(1+((v+p)/(1-pv))^2 ) (2)

u^'=1/(1-(v^2+2vp+p^2)/(1-2pv-〖pv〗^2 )) (2)

u^'=1/(1-2pv-〖pv〗^2-v^2+2vp+p^2 ) (2)

u^'=2/(1-〖pv〗^2-v^2+p^2 )

j) s= 1/√2 arc cos⁡〖√2t/(1-t^2 )〗

s^'=1/√(1-(√2t/(1-t^2 ))^2 ) (1/(√2t-t))

s^'=1/(√(2t/((1-2t^2-t^4 )) ) (1/(√2t-t))

s^'=1/(√2t/(√2t-t*√(1-2t^2-t^4 )) )

s^'=(√2t-t*√(1-2t^2-t^4 ))/√2t

s^'=√(-t+2t^3+t^5+2t-4t^3-2t^5 )/√2t

s^'=√(t-2t^3-t^5 )

〖k) y=(arccosx-arcsenx)〗^n

Y´=(n(arccosx-arcsenx))/(√(1-X^2 ) + √( 1-X^2 ))

Y´=(n(arccosx-arcsenx))/(2√(1-X^2 ) )

l) arccos(x-1)/√2

y´= - 1/√(1-〖((x+1)/√2)〗^2 ) * (√2-(x+1) 1/(2√2))/2

y´= - 1/√(1-〖((x+1)/√2)〗^2 ) * (√2- (x+1)/(2√2))/2

y´= - 1/√(1-〖((x+1)/√2)〗^2 ) *

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