What is the derivative of #(sinx)/(4+cosx)#?
1 Answer
Mar 17, 2018
Explanation:
#"differentiate using the "color(blue)"quotient rule"#
#"Given "y=(g(x))/(h(x))" then"#
#dy/dx=(h(x)g'(x)-g(x)h'(x))/(h(x))^2larrcolor(blue)"quotient rule"#
#g(x)=sinxrArrg'(x)=cosx#
#h(x)=4+cosxrArrh'(x)=-sinx#
#rArrd/dx((sinx)/(4+cosx))#
#=(cosx(4+cosx)+sin^2x)/(4+cosx)^2#
#=(4cosx+cos^2x+sin^2x)/(4+cosx)^2#
#=(4cosx+1)/(4+cosx)^2#
