differentiate: y = x^(1/2)*ln(x)?

Question

answer is (2+ln(x))/(2x^(1/2))

no idea how to arrive there. thanks

Answer 1

dy/dx = (1/2) x^(-1/2) ln x + (1/x) x^(1/2)
dy/dx = ln x / [ 2x^(1/2) ] + (x^(1/2) / x
dy/dx = [ x ln x + x^(1/2) 2x^(1/2) ] / [x(2x^(1/2)]
dy/dx = [ x lnx + 2x ] / [ ( x ) (2 x^(1/2)) ]
dy/dx = ( ln x + 2 ) / (2 √x)

Answer 2

If h(x) = f(x)*g(x) then d/dx h(x) = f'g + g'f

f'(x) = 1/2*x^(-1/2)

g'(x) = 1/x

dy/dx = 1/2*x^(-1/2) * ln(x) + 1/x * x^(1/2)

ln(x) / 2x^(1/2) + 2/2x^(1/2)

= (ln(x) + 2)/(2x^(1/2)

Answer 3

y'
= [x^(1/2)]'ln(x) + x^(1/2)[ln(x)]'
= lnx / (2x^(1/2)) + x^(1/2) / x
= (2+ln(x))/(2x^(1/2)), used product rule of differentiation.

Answer 4

y = x^(1/2)*ln(x)
dy/dx using the product rule:
dy/dx = d/dx(x^1/2) ln(x) + x^(1/2)d/dx(ln(x))
= (1/2)x^(-1/2) ln(x) + x^(1/2)(1/x)
= (1/(2x^(1/2) ) [ ln(x) + 2x ]
= ( ln(x) + 2x )/(2x^(1/2))

Answer 5

y = x^(1/2)*ln(x)
use product rule to differentiate,

dy/dx = x^(1/2)/x + 1/2x^(1/2) * ln(x)
dy/dx = 1/x^(1/2) + 1/2x^(1/2) * ln(x)
now take 2x^(1/2) as the LCM,
dy/dx = (2 + lnx)/(2x^(1/2))