Question

How do you differentiate `f(x)=(cotx)/(1+cotx)`?

Answer 1

`f'(x)=-(csc^2x)/(1+cotx)^2`

Explanation:

`f(x)=cotx/(1+cotx)`

`f'(x)=d/dx[cotx/(1+cotx)]`

`color(white)(f'(x))=((1+cotx)d/dx[cotx]-cotxd/dx[1+cotx])/(1+cotx)^2`

`color(white)(f'(x))=((1+cotx)(-csc^2x)-cotx(-csc^2x))/(1+cotx)^2`

`color(white)(f'(x))=(-csc^2x(1+cotx)+cotxcsc^2x)/(1+cotx)^2`

`color(white)(f'(x))=(-csc^2x(1+cotx-cotx))/(1+cotx)^2`

`color(white)(f'(x))=(-csc^2x(1))/(1+cotx)^2`

`color(white)(f'(x))=(-csc^2x)/(1+cotx)^2`

`color(white)(f'(x))=-(csc^2x)/(1+cotx)^2`

Answer 2

`f'(x)=-(csc^2x)/(1+cotx)^2`

Explanation:

`"differentiate using the "color(blue)"quotient rule"`

`"given "f(x)=(g(x))/(h(x))" then"`

`f'(x)=(h(x)g'(x)-g(x)h'(x))/(h(x))^2larrcolor(blue)"quotient rule"`

`g(x)=cotxrArrg'(x)=-csc^2x`

`h(x)=1+cotxrArrh'(x)=-csc^2x`

`rArrf'(x)=((1+cotx)(-csc^2x)+cotxcsc^2x)/(1+cotx)^2`

`color(white)(rArrf'(x))=(-csc^2xcancel(-cotxcsc^2c)cancel(+cotxcsc^2x))/(1+cotx)^2`

`=-(csc^2x)/(1+cotx)^2`