Question #91477

3 Answers
Somebody N.
Jan 7, 2018

#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)#

Nam D.
Jan 7, 2018

#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)#

Jim G.
Jan 7, 2018

#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#

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