Answer: f'(x) = x^6 (x-3)^8/(x^2+8)^2 * (6/x + 8/(x-3) - 4x/(x^2+8))
Step-by-step explanation:
Given: f(x) = x^6(x-3)^8/(x^2+8)^2
To find the derivative of f(x) = x^6(x-3)^8/(x^2+8)^2 using logarithmic differentiation, follow these steps:
1. Take the natural logarithm (ln) of both sides of the equation:
ln(f(x)) = ln(x^6(x-3)^8/(x^2+8)^2)
2. Apply the logarithmic properties to simplify the expression:
ln(f(x)) = 6ln(x) + 8ln(x-3) - 2ln(x^2+8)
3. Differentiate both sides of the equation with respect to x using the chain rule:
f'(x)/f(x) = 6/x + 8/(x-3) - 4x/(x^2+8)
4. Finally, multiply both sides by f(x) to find f'(x):
f'(x) = f(x) * (6/x + 8/(x-3) - 4x/(x^2+8))
f'(x) = x^6 (x-3)^8/(x^2+8)^2 * (6/x + 8/(x-3) - 4x/(x^2+8))
That's the derivative of the given function using logarithmic differentiation.