Question

How do you differentiate `f(x)=cotx` using the quotient rule?

Answer 1

`f'(x)=-csc^2x`.

Explanation:

`f(x)=cotx=cosx/sinx`

`:. f'(x)=d/dx(cosx/sinx)= (sinx(d/dxcosx)-cosxd/dx(sinx))/(sinx)^2`

`={sinx(-sinx)-cosx(cosx)}/sin^2x`

`=-(sin^2x+cos^2x)/sin^2x=-1/sin^2x`

`-csc^2x`.