Question

Question #d61e2

Answer 1

Given `y=csc x`, to find `(d^2y)/dx^2`

we know that `cscx = 1/sinx`
Hence, `y=1/sinx=(sinx)^-1`

Differentiating both sides we get
`dy/dx = -(sinx)^-2 cosx`

Differentiating again and using product rule
`(d^2y)/dx^2=d/dx[ -(sinx)^-2 cosx]`
`=>(d^2y)/dx^2=-[ (sinx)^-2 (-sinx)+cosx(-2(sinx)^-3cos x)]`
`=>(d^2y)/dx^2=[ (sinx)^-1 +2cos^2x(sinx)^-3]`
`=>(d^2y)/dx^2=cscx+2cot^2xcscx`
`=>(d^2y)/dx^2=cscx(1+2cot^2x)`