Respuesta :

MathPhys

Answer:

h'(x) = (-x² ln x + x² + 1) / (x (x² + 1)^(³/₂))

Step-by-step 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))

Wolfyy

Solution:

h(x) = ln(x)/√x^2+1

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

h(x) = ln(x) * (-1/2) * (x^2 + 1)^-3/2 * 2x + 1/x * (x^2 + 1)^-1/2

h(x) = -x ln(x) * (x^2 + 1)^-3/2 + 1/x * (x^2 + 1)^-1/2

h(x) = (x^2 + 1)^-3/2 * (-x ln(x) + 1/x * (x^2 + 1))

h(x) = -x^2ln(x)+x^2+1/(x(x^2+1)^3/2)

Best of Luck!

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