derivación mediante tablas
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Derivación mediante tablas
4. y=√2x2−2 x+1x
SoluciónMediante regla de división
d ( uv)
dx=−vu−uv´
v2
dydx=
x (√2 x2−2 x+1 )−x √2x2−2 x+1x2
=
x ( 4 x−2√2 x2−2 x+1 )−√2 x2−2 x+1
x2
dydx=
x (√2 x2−2 x+1 )−x √2x2−2 x+1x2
=
x ( 4 x−2√2 x2−2 x+1 )−√2 x2−2 x+1
x2
5: y=3√ x3+3 x2
xSolución
Mediante la regla de la división:d ( uv)
dx= vu−uv
v2
dydx
=
(3 x+6)3 3√(x3+3 x2)2
− 3√x3+3 x2
x2= x3+2 x2−x3−3 x2
x2 3√(x¿¿3+3 x2)= −x2
x2 3√(x3+3 x2)= 1
3√(x3+3 x2)2¿
6:y=(3 x2+4 x+8)√x−1SoluciónDerivada de un producto:d (uv )´=uv´ +vu´
y=(3x2+4 x+8 )√x−1→dydx
=(6 x+4 ) √x−1+ 3 x2+4 x+82√x−1
dydx
=12 x2+8x−12x−8+3x2+4 x+8
2√x−1= 15 x2
2√ x−1
7:y= xn
( x+1)n
Derivada de un cociente
dydx
=nxn−1(x+1)n−nxn(x+1)n+1
(x+1)2n
dydx
=nxn−1(x+1)n [x−1− (x−1 )−1]
(x+1)2n=nx
n ( x+1 )n(1+x−x)x (x+1)(x+1)2n
= nxn−1
(x+1)2n+1
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