Question

if y = xlogx then second order derivative is ?​

Answer 1

Answer:

hey there..!!!

Step-by-step explanation:

=> y = xlogx

then dy/dx = { logx.dx/dx + x.dlogx/dx }

= { logx.1 + x.1/x }

= { logx + 1}

now , d²y/dx² = {dlogx/dx + d1/dx}

= ( 1/x + 0)

= 1/x