What is the derivative of $y=\ln(\cot^{-1}\ x)$?

Question

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?

In the very first step there is an error. $-1$ is not an exponent, it is how to denote arccotangent. — vadim123, May 14 '15 at 14:29
I'm not really sure but we haven't really used cot^{-1} to denote 1/cot(x) so I assume it is arccot — , May 14 '15 at 15:37

score 1 · Accepted Answer · answered May 14 '15 at 14:38

$$\frac{d}{dx} \ln\left(\cot^{-1}(x)\right)$$

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 $-\frac{1}{x^2+1}$:

$$\frac{d}{dx} \ln\left(\cot^{-1}(x)\right)=\frac{\frac{d}{dx}(\cot^{-1}(x))}{\cot^{-1}(x)}=\frac{\frac{-1}{x^2+1}}{\cot{-1}(x)}=-\frac{1}{x^2\cot^{-1}(x)+\cot^{-1}(x)}$$

score 0 · Answer 2 · answered May 14 '15 at 14:33

0

by the chaine rule we obtain $$-\frac{1}{\left(x^2+1\right) \cot ^{-1}(x)}$$

answered May 14 '15 at 14:33

Dr. Sonnhard Graubner

95,283

score 0 · Answer 3 · answered May 14 '15 at 14:43

0

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

answered May 14 '15 at 14:43

David Holden

18,040

score 0 · Answer 4 · answered May 14 '15 at 14:51

$y = \ln(arccot x)$

$dy/dx = \frac{1}{arccotx} \frac{d}{dx} (arccotx)$

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{((tan z)(0) - (1)((\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:

enter image description here

Because: $\cot(theta) = x/1 = x$

$\to theta = arccot(x)$

Thus, $\sin(arccot(x)) = \frac{1}{\sqrt{x^2+1}}$

What is the derivative of $y=\ln(\cot^{-1}\ x)$?

4 Answers4