Question #d61e2

1 Answer
Feb 27, 2017

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)#