Use logarithmic differentiation to evaluate f'(x).
f(x)= ((x+1)^9) / (3x-6)^11
f'(x)=?
To use logarithmic differentiation, we take the natural logarithm of both sides of the equation:
ln(f(x)) = ln(((x+1)^9) / (3x-6)^11)
Next, we use the properties of logarithms to simplify the expression:
ln(f(x)) = ln((x+1)^9) - ln((3x-6)^11)
ln(f(x)) = 9ln(x+1) - 11ln(3x-6)
Now, we differentiate both sides of the equation with respect to x:
(d/dx) ln(f(x)) = (d/dx) (9ln(x+1)) - (d/dx) (11ln(3x-6))
Using the chain rule, we have:
f'(x)/f(x) = (9 * (1/(x+1))) - (11 * (1/(3x-6))) * (3)
Simplifying further:
f'(x)/f(x) = 9/(x+1) - 33/(3x-6)
Now, we can multiply both sides of the equation by f(x) to solve for f'(x):
f'(x) = f(x) * (9/(x+1) - 33/(3x-6))
Substituting f(x) back into the equation:
f'(x) = ((x+1)^9) / (3x-6)^11 * (9/(x+1) - 33/(3x-6))
Simplifying further if desired.