Find the derivative use logarithmic differentiation \( y=\sqrt[3]{\left(\frac{x^{2}+4}{x+1}\right)^{2}} \)

To differentiate the function \( y=\sqrt[3]{\left(\frac{x^{2}+4}{x+1}\right)^{2}} \) using logarithmic differentiation, we first take the natural logarithm of both sides: \[ \ln(y) = \ln\left(\sqrt[3]{\left(\frac{x^{2}+4}{x+1}\right)^{2}}\right) \] Using properties of logarithms, we can simplify the right side: \[ \ln(y) = \frac{1}{3} \ln\left(\left(\frac{x^{2}+4}{x+1}\right)^{2}\right) = \frac{2}{3} \ln\left(\frac{x^{2}+4}{x+1}\right) \] Now we apply the quotient rule for logarithms: \[ \ln(y) = \frac{2}{3} \left( \ln(x^2 + 4) - \ln(x + 1) \right) \] Next, we differentiate both sides with respect to \( x \): \[ \frac{1}{y} \frac{dy}{dx} = \frac{2}{3} \left( \frac{2x}{x^2 + 4} - \frac{1}{x + 1} \right) \] Now multiplying both sides by \( y \) and substituting back for \( y \): \[ \frac{dy}{dx} = y \cdot \frac{2}{3} \left( \frac{2x}{x^2 + 4} - \frac{1}{x + 1} \right) \] Substituting \( y = \sqrt[3]{\left(\frac{x^{2}+4}{x+1}\right)^{2}} \): \[ \frac{dy}{dx} = \sqrt[3]{\left(\frac{x^{2}+4}{x+1}\right)^{2}} \cdot \frac{2}{3} \left( \frac{2x}{x^2 + 4} - \frac{1}{x + 1} \right) \] This is the derivative of \( y \) using logarithmic differentiation.