Answer:
h'(x) = (-x² ln x + x² + 1) / (x (x² + 1)^(³/₂))
Explanation:
h(x) = ln x / √(x² + 1)
You can either use quotient rule, or you can rewrite using negative exponents and use product rule.
h(x) = (ln x) (x² + 1)^(-½)
h'(x) = (ln x) (-½) (x² + 1)^(-³/₂) (2x) + (1/x) (x² + 1)^(-½)
h'(x) = (-x ln x) (x² + 1)^(-³/₂) + (1/x) (x² + 1)^(-½)
h'(x) = (x² + 1)^(-³/₂) (-x ln x + (1/x) (x² + 1))
h'(x) = (1/x) (x² + 1)^(-³/₂) (-x² ln x + x² + 1)
h'(x) = (-x² ln x + x² + 1) / (x (x² + 1)^(³/₂))