Question

What is f '(x) for f(x) = secxcscx?

Answer 1

f(x)=secxcscx

let secx=g(x)

let cscx=h(x)

therefore, using leibnitz rule,

f'(x)=g'(x)*h(x)+h'(x)*g(x)

    Since derivative of secx=secxtanx

    and derivative of cscx=-cscxcotx,

f'(x)=secxtanxcscx -cscxcotxsecx

     =1/cosx*sinx/cosx*1/sinx - 1/sinx*cosx/sinx*1/cosx

     =1/cos^2x-1/sin^2x

     =(sin^2x-cos^2x)/sin^2xcos^2x / Option1