Questions: Use logarithmic differentiation to find dy/dx for the given function y. y=((x^3-1)^4 * sqrt(3x-1))/(x^2+4)

Solution Steps

To find the derivative of the given function using logarithmic differentiation, follow these steps:

1. Take the natural logarithm of both sides of the equation \( y = \frac{(x^3 - 1)^4 \cdot \sqrt{3x - 1}}{x^2 + 4} \).
2. Use the properties of logarithms to simplify the expression.
3. Differentiate both sides with respect to \( x \).
4. Solve for \( \frac{dy}{dx} \).

We start with the function given by \[ y = \frac{(x^3 - 1)^4 \cdot \sqrt{3x - 1}}{x^2 + 4}. \]

Taking the natural logarithm of both sides, we have: \[ \ln(y) = \ln\left(\frac{(x^3 - 1)^4 \cdot \sqrt{3x - 1}}{x^2 + 4}\right). \] Using properties of logarithms, this simplifies to: \[ \ln(y) = 4 \ln(x^3 - 1) + \frac{1}{2} \ln(3x - 1) - \ln(x^2 + 4). \]

Differentiating both sides with respect to \( x \) gives: \[ \frac{1}{y} \frac{dy}{dx} = \frac{12x^2}{x^3 - 1} + \frac{3}{2(3x - 1)} - \frac{2x}{x^2 + 4}. \] Multiplying through by \( y \) results in: \[ \frac{dy}{dx} = y \left( \frac{12x^2}{x^3 - 1} + \frac{3}{2(3x - 1)} - \frac{2x}{x^2 + 4} \right). \]

Substituting back for \( y \): \[ \frac{dy}{dx} = \frac{(x^3 - 1)^4 \cdot \sqrt{3x - 1}}{x^2 + 4} \left( \frac{12x^2}{x^3 - 1} + \frac{3}{2(3x - 1)} - \frac{2x}{x^2 + 4} \right). \]

Final Answer