Respuesta:
Y = (5∜(x+2x^3-4x^(-4)+2√x)/(2√x)
Y' = d/dx [ (5∜(x+2x^3-4x^(-4))+2√x)/(2√x) ]
Y' = ((5∜(x+2x^3-4x^(-4))+2√x)×d/dx[2√x])-(d/dx[ (5∜(x+2x^3-4x^(-4)+2√x)]×(2√x))/(2√x)^2
d/dx[ 2√x ] = 2×d/dx[√x ]
d/dx[ 2√x ] = 2(1/(2√x))
d/dx [ 2√x ] = 1/√x
d/dx [ (5∜(x+2x^3-4x^(-4))+2√x) ] = d/dx[
(5∜(x+2x^3-4x^(-4)) ]+d/dx[(2√x)]
d/dx[ (5∜(x+2x^3-4x^(-4)) ] = 5×d/dx[(∜(x+2x^3-4x^(-4))]
= 5×( 6x^7+x^5+16/((4x^2∜( 2x^7+x^5-4)^3)))
= (30x^7+5x^5+80/((4x^2∜( 2x^7+x^5-4)^3)))
Por ende :
d/dx [ 5∜(x+2x^3-4x^(-4)) ] = (30x^7+5x^5+80/((4x^2∜( 2x^7+x^5-4)^3)))
d/dx [ 2√(x) ] = 1/√(x)
Y' = ( 5∜(x+2x^3-4x^(-4))×1/√(x))-((30x^7+5x^5+80/((4x^2∜( 2x^7+x^5-4)^3)))(2√(x))/((2)^2(√(x))^2)
Y' = 60x^7√(x)+10x^5√(x)+160√(x)/((4x^2∜( 2x^7+x^5-4)^3))/(4×x)
Y ' = 60x^7√(x)+10x^5√(x)+160√(x)/((4x^2∜( 2x^7+x^5-4)^3/4x ====> Respuesta
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Respuesta:
Y = (5∜(x+2x^3-4x^(-4)+2√x)/(2√x)
Y' = d/dx [ (5∜(x+2x^3-4x^(-4))+2√x)/(2√x) ]
Y' = ((5∜(x+2x^3-4x^(-4))+2√x)×d/dx[2√x])-(d/dx[ (5∜(x+2x^3-4x^(-4)+2√x)]×(2√x))/(2√x)^2
d/dx[ 2√x ] = 2×d/dx[√x ]
d/dx[ 2√x ] = 2(1/(2√x))
d/dx [ 2√x ] = 1/√x
d/dx [ (5∜(x+2x^3-4x^(-4))+2√x) ] = d/dx[
(5∜(x+2x^3-4x^(-4)) ]+d/dx[(2√x)]
d/dx[ (5∜(x+2x^3-4x^(-4)) ] = 5×d/dx[(∜(x+2x^3-4x^(-4))]
= 5×( 6x^7+x^5+16/((4x^2∜( 2x^7+x^5-4)^3)))
= (30x^7+5x^5+80/((4x^2∜( 2x^7+x^5-4)^3)))
Por ende :
d/dx [ 5∜(x+2x^3-4x^(-4)) ] = (30x^7+5x^5+80/((4x^2∜( 2x^7+x^5-4)^3)))
d/dx [ 2√(x) ] = 1/√(x)
Y' = ( 5∜(x+2x^3-4x^(-4))×1/√(x))-((30x^7+5x^5+80/((4x^2∜( 2x^7+x^5-4)^3)))(2√(x))/((2)^2(√(x))^2)
Y' = 60x^7√(x)+10x^5√(x)+160√(x)/((4x^2∜( 2x^7+x^5-4)^3))/(4×x)
Y ' = 60x^7√(x)+10x^5√(x)+160√(x)/((4x^2∜( 2x^7+x^5-4)^3/4x ====> Respuesta
Explicación paso a paso: