How do you differentiate #y=1/(sinx+cosx)#?
1 Answer
Aug 16, 2017
Explanation:
#"we can use either the "color(blue)"quotient or chain rules"#
#y=1/(sinx+cosx)=(sinx+cosx)^-1#
#"differentiate using the "color(blue)"chain rule"#
#•color(white)(x)d/dx(f(g(x)))=f'(g(x)xxg'(x)#
#rArrdy/dx=-(sinx+cosx)^-2xxd/dx(sinx+cosx)#
#color(white)(rArrdy/dx)=-(sinx-cosx)/(sinx+cosx)^2#
#"differentiating 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))^2#
#g(x)=1rArrg'(x)=0#
#h(x)=sinx+cosxrArrh'(x)=cosx-sinx#
#rArrdy/dx=((sinx+cosx).0-(cosx-sinx))/(sinx+cosx)^2#
#color(white)(rArrdy/dx)=-(sinx-cosx)/(sinx+cosx)^2#