Question

derivada de y=sen^3 x^2 porfavor demen resolviendo es para mañana

Answer 1

Hallar la derivada- d/dx
y
=
sin
2
(
x
2
)
Diferencie usando la regla de la cadena, que establece que
d
d
x
[
f
(
g
(
x
)
)
]
es
f
'
(
g
(
x
)
)
g
'
(
x
)
donde
f
(
x
)
=
x
2
y
g
(
x
)
=
sin
(
x
2
)
.

2
sin
(
x
2
)
d
d
x
[
sin
(
x
2
)
]
Diferencie usando la regla de la cadena, que establece que
d
d
x
[
f
(
g
(
x
)
)
]
es
f
'
(
g
(
x
)
)
g
'
(
x
)
donde
f
(
x
)
=
sin
(
x
)
y
g
(
x
)
=
x
2
.

2
sin
(
x
2
)
(
cos
(
x
2
)
d
d
x
[
x
2
]
)
Diferenciar usando la regla de la potencia.
4
x
cos
(
x
2
)
sin
(
x
2
)

Answer 2

Solución:
$\textup{La expresi\'on es:}\\y=\sin^{3}(x^{2})=(\sin(x^{2}))^{3}\\\textup{Derivando se tiene:}\\y'=3(\sin(x^{2}))^{2}(2x)(1)\\y'=(3)(2x)(1)(\sin(x^{2}))^{2}\\y'=6x(\sin(x^{2}))^{2}\\y'=6x\sin^{2}(x^{2})\\\textup{o tambi\'en}\\y'=\frac{d}{dx}(\sin(x^{2}))^{3}=3(\sin(x^{2}))^{3-1}\frac{d}{dx}(x^{2})\\y'=3(\sin(x^{2}))^{2}(2x)\frac{d}{dx}(x)\\y'=3(\sin(x^{2}))^{2}(2x)(1)\\y=(\sin(x^{2}))^{2}(2x)\\y'=(2x)(3)(\sin(x^{2}))^{2}\\y'=6x(\sin(x^{2}))^{2}\\y'=6x\sin^{2}(x^{2})$
Saludos.