Question

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

Answer 1

`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`