$\frac{\partial \coth ^{-1}(x)}{\partial x}$ I am asked to find
$\dfrac{d\coth ^{-1}(x)}{dx}$
I rewrite it to become
$x=\dfrac{1}{\tan(y)}$
$\dfrac{\text{dx}}{\text{dy}}=-\dfrac{1}{\frac{\sec ^2(x)}{\tan ^2(x)}}=-\sin^2(x)$
However the answer should be 
$\dfrac{1}{1-x^2}$
What am I doing wrong?
 A: Let $y=\operatorname{\coth}^{-1} x$. Note that $\operatorname{coth}^{-1}$ is the inverse function of $\operatorname{\coth}$. Thus we have 
$$x=\operatorname{coth}(y)=\frac{\cosh y}{\sinh y}.$$
Differentiating, we obtain
$$\frac{dx}{dy}=\frac{\sin^2 y-\cosh^2 y}{\sinh^2 y}=1-\operatorname{coth}^2 y=1-x^2.$$
Thus (when $\frac{dx}{dy}$ is defined, and non-zero), we have $\frac{dy}{dx}=\frac{1}{\frac{dx}{dy}}=\frac{1}{1-x^2}$. 
A: You don't see the difference between $\coth^{-1}(x)$ and $1/\coth(x)$!
These are two different functions! Check with your textbook.
A: First note that
$$ \coth^{-1} x=\frac{1}{2}\ln\left(\frac{x+1}{x-1}\right), \mbox{for}\ |x|\gt 1$$
Then
$$ \frac{d}{dx}[\coth^{-1} x]=\frac{1}{2}\frac{d}{dx}\left[\ln\left(\frac{x+1}{x-1}\right)\right]$$
$$ =\frac{1}{2}\left[\frac{d}{dx}[\ln(x+1)]-\frac{d}{dx}[\ln(x-1)]\right] $$
$$ =\frac{1}{2}\left[\frac{1}{x+1}-\frac{1}{x-1}\right]=\frac{1}{2}\left[\frac{x-1-x-1}{(x+1)(x-1)}\right] $$
$$ =\frac{1}{2}\left[\frac{-2}{x^2-1}\right]=-\frac{1}{x^2-1} =\frac{1}{1-x^2}$$
