How do you differentiate #y=5x^3(sinx)^2#?
1 Answer
Aug 8, 2017
Explanation:
#"differentiate using the "color(blue)"product rule"#
#"given "y=g(x).h(x)" then "#
#dy/dx=g(x)h'(x)+h(x)g'(x)larr" product rule"#
#g(x)=5x^3rArrg'(x)=15x^2#
#h(x)=(sinx)^2rArrh'(x)=2sinxcosx=sin2x#
#rArrdy/dx=5x^3sin2x+15x^2sin^2x#
