### Solution 1

This solution involves using the chain rule (derivation made by Rigel Melaan).

Express acceleration as the ratio of velocity and time:

\(a(t)=\frac{\partial v}{\partial t}=\frac{\partial\left(\sqrt{\frac{m g}{k}} \tanh \left(\operatorname{arctanh}\left(v_0 \sqrt{\frac{k}{m g}}\right)-t \sqrt{\frac{k g}{m}}\right)\right)}{\partial t}\)

Factor out constants:

\(a(t)=\sqrt{\frac{m g}{k}} \times \frac{\partial\left(\tanh \left(\operatorname{arctanh}\left(v_0 \sqrt{\frac{k}{m g}}\right)-t \sqrt{\frac{k g}{m}}\right)\right)}{\partial t}\)

\(a(t)=\sqrt{\frac{m g}{k}} \times \frac{\partial\left(\tanh \left(\operatorname{arctanh}\left(v_0 \sqrt{\frac{k}{m g}}\right)-t \sqrt{\frac{k g}{m}}\right)\right)}{\partial t}\)

Consider the partial derivatives as a chain:

\(a(t)=\sqrt{\frac{m g}{k}} \times \frac{d \tanh \left(\operatorname{arctanh}\left(v_0 \sqrt{\frac{k}{m g}}\right)-t \sqrt{\frac{k g}{m}}\right)}{d\left(\operatorname{arctanh}\left(v_0 \sqrt{\frac{k}{m g}}\right)-t \sqrt{\frac{k g}{m}}\right)}\)

\(\times \frac{\partial\left(\operatorname{arctanh}\left(v_0 \sqrt{\frac{k}{m g}}\right)-t \sqrt{\frac{k g}{m}}\right)}{\partial t}\)

Using trigonometric identity for each derivative term in the chain rule:

\(\frac{d \tanh \left(\operatorname{arctanh}\left(v_0 \sqrt{\frac{k}{m g}}\right)-t \sqrt{\frac{k g}{m}}\right)}{d\left(\operatorname{arctanh}\left(v_0 \sqrt{\frac{k}{m g}}\right)-t \sqrt{\frac{k g}{m}}\right)}=\)

\(\operatorname{sech}^2\left(\operatorname{arctanh}\left(v_0 \sqrt{\frac{k}{m g}}\right)-t \sqrt{\frac{k g}{m}}\right)\)

\(\frac{\partial\left(\operatorname{arctanh}\left(v_0 \sqrt{\frac{k}{m g}}\right)-t \sqrt{\frac{k g}{m}}\right)}{\partial t}=\)

\(\frac{\partial\left(\operatorname{arctanh}\left(v_0 \sqrt{\frac{k}{m g}}\right)\right)}{\partial t}-\frac{\partial\left(t \sqrt{\frac{k g}{m}}\right)}{\partial t}\)

Simplify. Note that the derivative of a constant is zero:

\(\frac{\partial\left(\operatorname{arctanh}\left(v_0 \sqrt{\frac{k}{m g}}\right)-t \sqrt{\frac{k g}{m}}\right)}{\partial t}=\)

\(\frac{\partial\left(\operatorname{arctanh}\left(v_0 \sqrt{\frac{k}{m g}}\right)\right)}{\partial t}-\frac{\partial\left(t \sqrt{\frac{k g}{m}}\right)}{\partial t}\)

\(-\sqrt{\frac{k g}{m}} \frac{d(t)}{d t}=-\sqrt{\frac{k g}{m}}\)

Apply the chain rule: the derivatives of the original function is the product of individual derivatives:

\(a(t)=\sqrt{\frac{m g}{k}} \times \operatorname{sech}^2\left(\operatorname{arctanh}\left(v_0 \sqrt{\frac{k}{m g}}\right)-t \sqrt{\frac{k g}{m}}\right) \times-\sqrt{\frac{k g}{m}}\)

Simplify to get the acceleration-time model equation for free-falling bodies with air resistance:

\(a(t)=-g \operatorname{sech}^2\left(\operatorname{arctanh}\left(v_0 \sqrt{\frac{k}{m g}}\right)-t \sqrt{\frac{k g}{m}}\right)\)