Question

What is the second order derivative of y=x²lnx?
Differentiate cosx/1+cosx.

Answer 1

Answer:

solved

Step-by-step explanation:

i) y=x²lnx

y' = x².1/x  + 2x ln x = 2xlnx + x

y'' = 2x/x + 2lnx + 1 = 3 + 2lnx

ii) Differentiate cosx/1+cosx.

y = cosx/1+cosx

y' = {-sinx(1+cosx) - cosx(1-sinx)} ÷ (1+cosx)² =   - { sinx + cosx} ÷ (1+cosx)²