Question

What is the second derivative of `f(x)=-sec2x-cotx`?

Answer 1

`f''(x)=-4sec2x{sec^2(2x)+tan^2(2x)}-2csc^2xcotx.`

Explanation:

`f(x)=-sec2x-cotx`
`rArr f'(x)=(-sec2x)'-(cotx)'=-sec2x*tan2x*d/dx(2x)-(-csc^2x)`
`=-2sec2x*tan2x+csc^2x`
`rArr f''(x)={f'(x)}'=(-2sec2xtan2x)'+{(cscx)^2}'`.
`=-2{sec2x*(tan2x)'+tan2x*(sec2x)'}+2cscx*(cscx)'`, ......[Product Rule & Chain Rule]
`=-2[sec2x*{sec^2(2x)*2}+tan2x{sec2x*tan2x*2}]+2cscx(-cscx*cotx),`
`=-2{2sec^3(2x)+2sec2x*tan^2(2x)}-2csc^2xcotx,`
`=-4sec2x{sec^2(2x)+tan^2(2x)}-2csc^2xcotx.`