Community Answer

【General guidance】The answer provided below has been developed in a clear step by step manner.Step1/2Ans.To show the given statement by mathematical induction, we need to first verify the base case, and then show that if the statement holds for some integer n, it also holds for n+1.Base case:We want to show that the statement holds for n=1, i.e., that the first derivative of f(x) is given by:f'(x) = (-1)^2 * 3^1 * 2!/ (3x+5)^1 = -6/(3x+5)Now, f(x) = ln(3x+5), and using the chain rule, we have:f'(x) = 1/(3x+5) * d/dx(3x+5) = 1/(3x+5) * 3 = 3/(3x+5)Multiplying by (-2) in the numerator and denominator givesf'(x) = (-6)/(3x+5), which is the expression we wanted to show for the base case.Inductive step:Assume that the statement holds for some integer k, i.e., that the kth derivative of f(x) is given by:f^(k)(x) = (-1)^(k+1) * 3^k * (k+1)! / (3x+5)^kWe want to show that this implies the statement for k+1, i.e., that the (k+1)th derivative of f(x) is given by:f^(k+1)(x) = (-1)^(k+2) * 3^(k+1) * (k+2)! / ... See the full answer