Problem

(a) Starting from the definitions of $\tanh$ and $\operatorname{sech}$ in terms of exponentials, prove that

$$1 - \tanh^2 u = \operatorname{sech}^2 u.$$

(b) Show that $\frac{\mathrm{d}}{\mathrm{d}t}(\operatorname{sech}^{-1} t) = -\frac{1}{t\sqrt{1 - t^2}}$.

It is given that

$$\begin{align*} x =&\, \tanh^{-1} t \quad\text{and}\quad y = t\operatorname{sech}^{-1} t, \quad \text{for } 0 < t < 1. \end{align*}$$

(c) Show that $\frac{\mathrm{d}y}{\mathrm{d}x} = -\sqrt{1 - t^2} + (1 - t^2)\operatorname{sech}^{-1} t$.

(d) Find $\frac{\mathrm{d}^2y}{\mathrm{d}x^2}$ in terms of $t$.