A quick ChatGPT produce this tanh derivative calculation:

As

\tanh(x) = \frac{e^x - e^{-x}}{e^x + e^{-x}}

Let y = \tanh(x). So,

y = \frac{e^x - e^{-x}}{e^x + e^{-x}}

Now, denote u = e^x and v = e^{-x}. Thus,

y = \frac{u - v}{u + v}

Then, using the quotient rule, the derivative of y with respect to x can be calculated as follows:

\frac{dy}{dx} = \frac{(v \cdot u' - u \cdot v')}{(v + u)^2}

Where ( u') and ( v' ) are the derivatives of ( u ) and ( v ) with respect to ( x), respectively.

u' = e^x

v' = -e^{-x}

Substitute these into the formula:

\frac{dy}{dx} = \frac{(e^{-x} \cdot e^x - e^x \cdot (-e^{-x}))}{(e^{-x} + e^x)^2}

\frac{dy}{dx} = \frac{(e^{x-x} + e^{x-x})}{(e^x + e^{-x})^2}

As e^0 = 1

\frac{dy}{dx} = \frac{1+1}{(e^x + e^{-x})^2}

\frac{dy}{dx} = \frac{2}{(e^x + e^{-x})^2}

Since {e^x + e^{-x}} = 2cosh(x), where cosh(x) is the hyperbolic cosine function, we can substitute this back:

\frac{dy}{dx} = \frac{2}{(2\cosh(x))^2}

\frac{dy}{dx} = \frac{1}{\cosh^2(x)}

\frac{dy}{dx} = \text{sech}^2(x)

And \text{sech}^2(x) is equal to 1 - \text{tanh}^2(x).