What is the derivative of $y=\ln(\cot^{-1}\ x)$? What is the derivative of $y$? 
$$y=\ln(\cot^{-1}\ x)$$
If I take out the exponent $-1$, so
$$y=-\ln(\cot\ x)$$
Getting the derivative would be
$$dy=-\frac{1}{\cot\ x}\ (-\csc^{2}\ x)\ dx$$ 
$$dy=\frac{1}{\sin\ x\ \cos\ x}\ dx$$
But if we don't take out the exponent, the derivative would be
$$dy=\frac{1}{\cot^{-1}\ x}\ (\frac{-1}{1+x^{2}})\ dx$$ 
$$dy=-\frac{1}{(\cot^{-1}\ x)\ (1+x^{2})}\ dx$$ 
But if I assign a value for $x$, I get different values.
Which is the correct derivative? Or did I miss a step in derivation? Did I just possibly plugged in wrong values that's why I get different values for the two derivatives?
 A: $$\frac{d}{dx} \ln\left(\cot^{-1}(x)\right)$$
Using the chain rule, $\frac{d}{dx} \ln\left(\cot^{-1}(x)\right)=\frac{d\ln(u)}{du}\frac{du}{dx}$ where $u=\cot^{-1}(x)$ and $\frac{d}{du}(\ln(u))=\frac{1}{u}$:
$$\frac{d}{dx} \ln\left(\cot^{-1}(x)\right)=\frac{\frac{d}{dx}(\cot^{-1}(x))}{\cot^{-1}(x)}$$
The derivative of $\cot^{-1}(x)$ is $-\frac{1}{x^2+1}$:
$$\frac{d}{dx} \ln\left(\cot^{-1}(x)\right)=\frac{\frac{d}{dx}(\cot^{-1}(x))}{\cot^{-1}(x)}=\frac{\frac{-1}{x^2+1}}{\cot{-1}(x)}=-\frac{1}{x^2\cot^{-1}(x)+\cot^{-1}(x)}$$
A: by the chaine rule we obtain
$$-\frac{1}{\left(x^2+1\right) \cot ^{-1}(x)}$$
A: $$
e^y = \cot^{-1}x \\
e^y dy = \frac{-dx}{1+x^2} \\
\frac{dy}{dx} =  \frac{-1}{(1+x^2)e^y} = \frac{-1}{(1+x^2)\cot^{-1}x}
$$
A: $y = \ln(arccot x)$
$dy/dx = \frac{1}{arccotx} \frac{d}{dx} (arccotx)$
Use implicit differentiation to get $\frac{d}{dx} (arccotx)$:
$z = arccotx$
$\cot z = x$
$\frac{d}{dx} \cot z = \frac{d}{dx} x$
$\frac{d}{dx} \frac{1}{\tan z} = 1$
$\frac{d}{dx} \frac{1}{\tan z} = 1$
$\frac{((tan z)(0) - (1)((\sec^2 z) z')}{\tan^2 z} = 1 $
$\frac{(- (\sec^2 z) z')}{\tan^2 z} = 1 $
$\frac{(- (\sec^2 z) z')}{\tan^2 z} = 1 $
$(- (\sec^2 z) z') = \tan^2 z$
$z' = \frac{-\tan^2 z}{\sec^2 z}$
$z' = -\sin^2 z$
$z' = -(\sin z)^2$
$\frac{d}{dx} (arccotx) = -(\sin z)^2$
where $\sin z$
$= \sin(arccot(x))$
Now, "$arccot(x)$" is the theta in the following right triangle:

Because: $\cot(theta) = x/1 = x$
$\to theta = arccot(x)$
Thus, $\sin(arccot(x)) = \frac{1}{\sqrt{x^2+1}}$
