Find y' and y''. y = sqrt x ln(x)

Question

Please help me

Answer 1

y'=(lnx+2)/(2√x)        -------lnx=log_ex

y''=[(lnx+2)/(2√x)]'       --------------      use quotient rule
=1/2{(1/x +0)*√(x)+(lnx +2)*(-1/2*x^-3/2)}

=(1/2x√x) -(lnx)/4x√x  -((1/2x√x) )

=lnx/4x√x

Answer 2

y=√x.lnx = x^1/2.log_ex

Hence y' =1/2 x^-1/2.log_ex+x^1/2.1/x ... differentiating both sides w.r.t x

 

             =[log_ex/(2√x) ] + 1/√x ..... By adjusting the powers using principles of indices

             =[log_ex+2]/(2√x)