Error in logarithmic differentiation of $R(s)=s^{\ln s}$

Question

I was trying to solve for the derivative of $R(s) = S^{ln(s)}$. I understand that there is a much simpler way to do it through a single use of the chain rule, but I wanted to see if I could figure out how to solve for the derivative of a logarithm, with any base, of any function. What I can not understand is why I am getting the wrong answer.

$R(s) = s^{\ln(s)}$

$\log_s R(s) = \log_s s^{\ln(s)}$

Logorithmic change of base:

$\frac{\ln R(s)}{\ln s} = \ln(s)$

Take the derivative of both sides and apply the quotient rule of derivatives:

$\frac{ \ln(R(s))^{\prime} \ln(s) - \ln(s)^{\prime} \ln(s)}{(\ln(s))^2} = \frac{1}{s}\qquad\qquad (*)$

$\frac{\frac{R^{\prime}(s)}{R(s)} \ln(s) - \frac{\ln(s)}{s}}{(\ln(s))^2} = \frac{1}{s}$

$\frac{\ln(s)^2}{s} = \frac{R^\prime (s)}{R(s)} \ln(s)-\frac{\ln(s)}{s}$

$\ln(s)^2 +\frac{ln(s)}{s} = \frac{R^\prime (s) \ln(s)}{R(s)}\qquad\qquad (**)$

$\frac{s^{\ln(s)}}{ln(s)} (\ln(s)^2 + \frac{\ln(s)}{s}) = R^\prime $

But the correct answer is: $\frac{2 \ln(s)}{s} s^{\ln(s)} = \frac{dR(x)}{dx}$

Where do I make my mistake?

Answer 1

1. The equation (*) and the next after: the mistake is marked in red $$ \frac{\frac{R^{\prime}(s)}{R(s)} \ln(s) - \frac{\color{red}{\ln(s)}}{s}}{(\ln(s))^2} = \frac{1}{s}. $$ It must be $\color{green}{\ln R(s)}$.
2. The equation (**): $\frac{1}{s}$ is missing in the first term.