How do you differentiate #f(x)=sin^3xcosx#?

1 Answer
Jim G.
Jun 28, 2017

#f'(x)=3sin^2xcos^2x-sin^4x#

Explanation:

#"differentiate using the "color(blue)"product rule"#

#"given " f(x)=g(x).h(x)" then"#

#f'(x)=g(x)h'(x)+h(x)g'(x)larr" product rule"#

#g(x)=sin^3x=(sinx)^3#

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

#g'(x)=3(sinx)^2xxd/dx(sinx)=3sin^2xcosx#

#h(x)=cosxrArrh'(x)=-sinx#

#rArrf'(x)=sin^3x(-sinx)+3sin^2xcosx(cosx)#

#color(white)(rArrf'(x))=3sin^2cos^2x-sin^4x#

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