Use logarithmic differentiation to find dy/dx

Question

Use logarithmic differentiation to find dy/dx

y = x√x²+1 note x > 0

Answer 1

y = x √(x² + 1)

lny = ln [ x √(x² + 1) ]

lny = lnx + (1/2)ln(x² + 1)

y' / y = 1/x + (1/2)(2x / (x²+1))

y' / y = 1/x + x/(x² + 1) = (2x²+1) / [x(x²+1)]

y' = y(2x²+1) / [x(x²+1)] = x√(x²+1)(2x²+1) / [x(x²+1)] = (2x²+1) / √(x²+1)

Answer 2

y= xsqr(x^2+1)

use the product rule and simplify

y'= sqr(x^2+1) + x[sqr(x^2+1)]'

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

y'= dy/dx = (2x^2+1)/sqr(x^2+1)