Is my proof correct for $\tanh{(\tau+\epsilon)}=\tanh{(\tau)}+\epsilon(1-\tanh^2{\zeta})$ correct?

Question

I am trying to prove an equality using Mean Value Theorem, I wish to know if my reasoning is correct, or if someone can improve upon the proof. The expression I am trying to prove is: \begin{equation} \tanh{(\tau+\epsilon)}=\tanh{(\tau)}+\epsilon(1-\tanh^2{\zeta}) \end{equation} where, $0\leq \zeta \leq \tau$. My proof: \begin{equation} \tanh{(\tau+\epsilon)}=\int_0^{\tau+\epsilon}(1-\tanh^2{x})dx \end{equation}, and Using the MVT, the RHS of equation above, can be written as: \begin{equation} \begin{split} \int_0^{\tau+\epsilon}(1-\tanh^2{x})dx&=(\tau+\epsilon)(1-\tanh^2{\zeta})\\ &=\tau(1-\tanh^2{\zeta})+\epsilon(1-\tanh^2{\zeta}) \end{split} \end{equation}. where $0\leq \zeta \leq \tau+\epsilon$ . Now, \begin{equation} (1-\tanh^2{\zeta})=\frac{d(\tanh{x})}{dx}\Big|_\zeta \end{equation}, Now, using the derivative version of the MVT on the RHS of equation above, one gets: \begin{equation} \frac{d(\tanh{x})}{dx}\Big|_\zeta=\frac{\tanh{(\tau})-0}{\tau-0} \end{equation}. Under the assumption, $0\leq \zeta\leq \tau$. Using all the above, it becomes, \begin{equation} \tanh{(\tau+\epsilon)}=\tanh{(\tau)}+\epsilon(1-\tanh^2{\zeta}) \end{equation}. Is my proof correct? Please comment! Thanks for your time and consideration!

Answer 1

The result is more immediate. By MVT, for some $\zeta \in (\tau,\tau +\varepsilon)$ $$\frac{\tanh (\tau + \varepsilon) - \tanh (\tau)}{\varepsilon} = \tanh '(\zeta) = 1-\tanh ^2 (\zeta) $$ I am not sure how we were supposed to pick a suitable $0\leq \zeta\leq\tau$, however. In fact, I'm quite sure you won't be able to do that. For a sufficiently large $n = n(\varepsilon)$ $$\varepsilon > \tanh (n+1) - \tanh (n) = 1-\tanh ^2(\zeta) \geq 1 - \tanh (\zeta) $$ yielding $\tanh (\zeta) + \varepsilon >1$. If $\varepsilon >0$ is small, then no $0\leq \zeta\leq n$ would satisfy that demand.