Final answer:
Logarithmic differentiation involves taking the natural logarithm on both sides, using properties of logarithms to simplify the expression, differentiating both sides with respect to x, applying the chain rule, solving for dy/dx, and then substituting the original function back into the equation to get the final derivative.
Step-by-step explanation:
To use logarithmic differentiation to find dy/dx for the function y=(3x^4-2)^5/(3x^3+4)^2, we first apply the properties of logarithms to take the natural logarithm on both sides:
ln(y) = ln((3x^4-2)^5/(3x^3+4)^2)
Using the properties of logarithms, we can write it as:
ln(y) = 5*ln(3x^4-2) - 2*ln(3x^3+4)
Now, we differentiate both sides with respect to x.
d/dx [ln(y)] = d/dx [5*ln(3x^4-2)] - d/dx [2*ln(3x^3+4)]
We apply the chain rule to differentiate the logarithmic terms and the quotient rule to differentiate the left side:
(1/y)(dy/dx) = (5/(3x^4-2))(12x^3) - (2/(3x^3+4))(9x^2)
Now solve for dy/dx:
dy/dx = y[(5*12x^3)/(3x^4-2) - (2*9x^2)/(3x^3+4)]
Finally, substitute the original y back into the equation:
dy/dx = ((3x^4-2)^5/(3x^3+4)^2)[(60x^3)/(3x^4-2) - (18x^2)/(3x^3+4)]