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

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: $$$$\tanh{(\tau+\epsilon)}=\tanh{(\tau)}+\epsilon(1-\tanh^2{\zeta})$$$$ where, $$0\leq \zeta \leq \tau$$. My proof: $$$$\tanh{(\tau+\epsilon)}=\int_0^{\tau+\epsilon}(1-\tanh^2{x})dx$$$$, and Using the MVT, the RHS of equation above, can be written as: $$$$\begin{split} \int_0^{\tau+\epsilon}(1-\tanh^2{x})dx&=(\tau+\epsilon)(1-\tanh^2{\zeta})\\ &=\tau(1-\tanh^2{\zeta})+\epsilon(1-\tanh^2{\zeta}) \end{split}$$$$. where $$0\leq \zeta \leq \tau+\epsilon$$ . Now, $$$$(1-\tanh^2{\zeta})=\frac{d(\tanh{x})}{dx}\Big|_\zeta$$$$, Now, using the derivative version of the MVT on the RHS of equation above, one gets: $$$$\frac{d(\tanh{x})}{dx}\Big|_\zeta=\frac{\tanh{(\tau})-0}{\tau-0}$$$$. Under the assumption, $$0\leq \zeta\leq \tau$$. Using all the above, it becomes, $$$$\tanh{(\tau+\epsilon)}=\tanh{(\tau)}+\epsilon(1-\tanh^2{\zeta})$$$$. Is my proof correct? Please comment! Thanks for your time and consideration!

• How do you justify using the same $\zeta$ in both applications of the MVT? Aug 24, 2019 at 11:41
• You obtained some $0\leq \zeta _1\leq\tau +\varepsilon$ and $0\leq \zeta _2 \leq\tau$. Do you claim they are equal? Aug 24, 2019 at 12:19

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.