Question

Question #91477

Answer 1

`3sin^2(x)cos(x)`

Explanation:

`sin^3(x)<=>(sinx)^3`

Using the chain rule:

`dy/dx=dy/(du)*(du)/dx`

Let `u=sin(x)`

`dy/dx(sin^3(x))=dy/(du)(u)^3*(du)/dx(u)`

`color(white)(88888888888)=3u^2*cos(x)`

`u=sin(x)`

`color(white)(88888888888)=3sin^2(x)cos(x)`

Answer 2

`3sin^2(x)cos(x)`

Explanation:

Using the chain rule,

`(df(u))/dx=(df)/(du)*(du)/(dx)`

Let `u=sin(x)`

`(du)/(dx)=cos(x)`

`f=u^3`

`(df)/(du)=3u^2`

`:.(df(u))/dx=3u^2cos(x)`

Substitute `u=sin(x)` back, we get

`:.(df(u))/dx=3sin^2(x)cos(x)`

Answer 3

`3sin^2xcosx`

Explanation:

`"differentiate using the "color(blue)"chain rule"`

`"given "y=f(g(x))" then"`

`dy/dx=f'(g(x))xxg'(x)larrcolor(blue)"chain rule"`

`"express "sin^3x=(sinx)^3`

`rArrd/dx((sinx)^3)`

`=3(sinx)^2xxd/dx(sinx)=3sin^2xcosx`