Simplify the derivative of hyperbolic tangent using $y$

Suppose $$y=\tanh(z)=\frac{\exp(z) − \exp(−z)}{\exp(z) + \exp(−z)}=\frac{\exp(2z) − 1}{\exp(2z) + 1}$$

$$u = \exp(2 z) -1$$ $$v=\exp(2 z)+1$$ $$\frac{\partial y}{\partial z} = \frac{\partial}{\partial z} \left(\frac{u}{v} \right) = \frac{v \frac{\partial u}{\partial a^t_j} - u \frac{\partial v}{\partial z}}{v^2} = \frac{4 \exp (2 z)}{(\exp(2 z)+1)^2}$$

I would be interested in expressing $$\frac{\partial y}{ \partial z}$$ in function of $$y$$ directly. I got $$\frac{\partial y}{ \partial z} = \frac{4}{\exp(2z)+1} \left( y + \frac{1}{\exp(2z) +1}\right)$$

Is there a way to simplify that?

• Given your definition of $y$, you have $z = \tanh^{-1}$ where it is defined. Given $\partial y/ \partial z$ as you've calculated, and some trig identities for inverse arctan, can you see how to simplify? – Derek Allums Oct 23 at 15:12

Suppose $$y=\tanh(z)$$ and
$$x=\frac{\partial y}{ \partial z} = \text {sech}^2 (z)$$
$$x =1 - y^2$$