Show that ${d\over dx}\tanh^{-1}x = {1 \over 1-x^2}$ 
Claim: 
  $${d\over dx}\tanh^{-1}x = {1 \over 1-x^2}$$

My work.
Let $y = \tanh^{-1}x$. Then
$$\tanh y =x \Rightarrow {d \tanh y \over dy}\cdot {dy\over dx} = 1$$
Since $\tanh y = \dfrac{e^y - e^{-y}}{e^y + e^{-y}}$ then
$${d \tanh y \over dy} = d {(e^y - e^{-y})(e^y + e^{-y})^{-1} \over dy}=1 - (e^y - e^{-y})(e^y + e^{-y})^{-2}(e^y - e^{-y})= 0$$
which means $0 \cdot {dy\over dx} = 1$
I think I did something wrong but I cannot find which part it is.
Please, give me some hints or advice.
 A: The problem is
$$
(e^y - e^{-y})(e^y + e^{-y})^{-2}(e^y - e^{-y}) \neq 1
$$
An alternative avoiding exponentials is
$$
\frac{d \tanh(y)}{dy}=\frac{d \sinh(y)/\cosh(y)}{dy}=\frac{\cosh^2(y)-\sinh^2(y)}{\cosh^2(y)}=1-\tanh^2(y)=1-x^2
$$
A: You approach is correct but
$${d \tanh y \over dy} = {d \over dy} \left(\frac{e^y - e^{-y}}{e^y + e^{-y}}\right)=\frac{(e^y + e^{-y})(e^y + e^{-y})-(e^y - e^{-y})(e^y - e^{-y})}{(e^y + e^{-y})^2}\\=\frac{4}{(e^y + e^{-y})^2}=\frac{1}{\cosh^2 y}.$$
Now note that
$$\tanh^2 y=\frac{\sinh^2 y}{\cosh^2 y}=\frac{\cosh^2 y-1}{\cosh^2 y}=1-\frac{1}{\cosh^2 y}$$
where we used the identity $\cosh^2 y-\sinh^2 y=1$.
Can you take it from here?
A: Notice that
\begin{align}
(e^y-e^{-y})(e^y+e^{-y})^2(e^y-e^{-y}) &= \left( \frac{e^y+e^{-y}}{e^y-e^{-y}}\right)^2 \\
&= \frac{1}{\tanh^2y}
\end{align}
Then proceed from there, you have the correct method, and have tried to show from a greater detail than relying on standard derivatives and for that you should be applauded.
A: I allow myself to post this answer that I like and I don't think this has already been mentionned:
We know that 
$$\tanh(\tanh^{-1}(x)) = x$$
Taking the derivative on both sides :
$$
\tanh'(\tanh^{-1}(x)) \cdot \tanh^{-1}(x)' = 1 \Leftrightarrow \tanh^{-1}(x)' = \frac{1}{\tanh'(\tanh^{-1}(x)) }
$$
Knowing that $\tanh'(x) = 1-\tanh(x)^2$ we get :
$$
\tanh^{-1}(x)' = \frac{1}{\tanh'(\tanh^{-1}(x)) } = \frac{1}{1-\tanh^2(\tanh^{-1}(x))} = \frac{1}{1-x^2}
$$
