If \( f(x)=(6 x-5)^{3} \cdot\left(8 x^{2}+4\right)^{3} \), find \( f^{\prime}(x) \) using logarithmic differentiation. Submit Question

To differentiate the function \( f(x) = (6x - 5)^3 \cdot (8x^2 + 4)^3 \) using logarithmic differentiation, we first take the natural logarithm of both sides: \[ \ln(f(x)) = \ln((6x - 5)^3) + \ln((8x^2 + 4)^3) \] Using the properties of logarithms, this simplifies to: \[ \ln(f(x)) = 3 \ln(6x - 5) + 3 \ln(8x^2 + 4) \] Next, we differentiate both sides with respect to \( x \): \[ \frac{f'(x)}{f(x)} = 3 \cdot \frac{1}{6x - 5} \cdot 6 + 3 \cdot \frac{1}{8x^2 + 4} \cdot (16x) \] This gives us: \[ \frac{f'(x)}{f(x)} = \frac{18}{6x - 5} + \frac{48x}{8x^2 + 4} \] Now, multiply through by \( f(x) \) to isolate \( f'(x) \): \[ f'(x) = f(x) \left( \frac{18}{6x - 5} + \frac{48x}{8x^2 + 4} \right) \] Replacing \( f(x) \) back in, we have: \[ f'(x) = (6x - 5)^3 (8x^2 + 4)^3 \left( \frac{18}{6x - 5} + \frac{48x}{8x^2 + 4} \right) \] Now, this is your \( f'(x) \) with all components in place! You can further simplify to get a cleaner expression if needed.