\displaystyle{f}'{\left({x}\right)}=\frac{{2}}{{{x}{\left({1}-{\ln{{x}}}\right)}^{{2}}}} Explanation: Let \displaystyle{t}={\ln{{x}}} Then \displaystyle\frac{{d}}{{\left.{d}{x}\right.}}{f{{\left({x}\right)}}}=\frac{{\left.{d}{t}\right.}}{{\left.{d}{x}\right.}}\frac{{d}}{{\left.{d}{t}\right.}}\frac{{{1}+{t}}}{{{1}-{t}}} ...

\displaystyle\frac{{d}}{{\left.{d}{x}\right.}}\frac{{{1}-{\ln{{x}}}}}{{{1}+{\ln{{x}}}}}=-\frac{{{2}}}{{{x}{\left({1}+{\ln{{x}}}\right)}^{{2}}}} Explanation: Using the quotient rule, we have: \displaystyle\frac{{d}}{{\left.{d}{x}\right.}}\frac{{{1}-{\ln{{x}}}}}{{{1}+{\ln{{x}}}}}=\frac{{{\left({1}+{\ln{{x}}}\right)}{\left(\frac{{d}}{{\left.{d}{x}\right.}}{\left({1}-{\ln{{x}}}\right)}\right)}-{\left({1}-{\ln{{x}}}\right)}{\left(\frac{{d}}{{\left.{d}{x}\right.}}{\left({1}+{\ln{{x}}}\right)}\right)}}}{{\left({1}+{\ln{{x}}}\right)}^{{2}}} ...

Let g(x)=f(x+1) so that g^{(n)}(0)=f^{(n)}(1). Then since \log(1+x) = \int dx/(1+x) =\sum_{n=0}^\infty {(-1)^n x^{n+1}/ (n+1)} g(x)={\log(1+x)\over1+x}=\sum_{n=0}^\infty(-1)^nx^n \sum_{n=0}^\infty {(-1)^n x^{n+1} \over n+1}=\sum_{n=0}^\infty(-1)^nx^n \sum_{m=1}^\infty {(-1)^{m-1} x^{m} \over m} = \sum_{n=1}^\infty\left(\sum_{k=1}^n{(-1)^{n-1} \over k}\right) x^n ...

Write your derivative as \ln{\left(1+\frac1x\right)}-\frac{\frac1x}{1+\frac1x} and use the inequality \ln{(1+x)}>\frac{x}{1+x} for 0<x<1 . Edit: By definition \ln{(1+x)}:=\int_{1}^{1+x}\frac1tdt>\frac1{1+x}\int_{1}^{1+x}dt=\frac{x}{1+x} ...

See edit below I don't think the solution should be independent of c. Maybe you typed into Wolfram a^x+b^x=0 In that case, it's not too hard to see where the solution comes from. Move b^x ...

You approach is good, but there is another which is worthy of consideration. Define h : (0,\infty) \rightarrow \mathbb{R} by \begin{equation} h(x) = x \text{ln}(x) - x + 1. \end{equation} and show ...