How do you differentiate #y=1/(sinx+cosx)#?

1 Answer
Aug 16, 2017

#dy/dx=-(sinx-cosx)/(sinx+cosx)^2#

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#