Use logarithmic differentiation to determine $$ y^{\prime} $$ for the equation $$ y=\sqrt[3]{(2 x-3)(x-6)(4 x+5)} $$

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Given
$$
y=\sqrt[3]{(2x-3)(x-6)(4x+5)}=\bigl((2x-3)(x-6)(4x+5)\bigr)^{\frac{1}{3}},
$$
take natural logs of both sides:
$$
\ln y=\frac{1}{3}\Bigl[\ln(2x-3)+\ln(x-6)+\ln(4x+5)\Bigr].
$$

Differentiate both sides with respect to $$ x $$. Using the chain rule on the left and the derivative of logarithms on the right:
$$
\frac{y'}{y}=\frac{1}{3}\left[\frac{2}{2x-3}+\frac{1}{x-6}+\frac{4}{4x+5}\right].
$$

Solving for $$ y' $$ gives:
$$
y'=y\cdot\frac{1}{3}\left[\frac{2}{2x-3}+\frac{1}{x-6}+\frac{4}{4x+5}\right].
$$

Substitute back the expression for $$ y $$:
$$
y'=\sqrt[3]{(2x-3)(x-6)(4x+5)}\cdot\frac{1}{3}\left[\frac{2}{2x-3}+\frac{1}{x-6}+\frac{4}{4x+5}\right].
$$
To find $$ y' $$ using logarithmic differentiation for $$ y = \sqrt[3]{(2x-3)(x-6)(4x+5)} $$, take the natural logarithm of both sides, differentiate implicitly, and solve for $$ y' $$. The final expression for $$ y' $$ is: $$ y' = \sqrt[3]{(2x-3)(x-6)(4x+5)} \cdot \frac{1}{3} \left( \frac{2}{2x-3} + \frac{1}{x-6} + \frac{4}{4x+5} \right) $$

Use logarithmic differentiation to determine for the equation

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