Johnsy's question about finding a derivative via Facebook

[Prove It]

How do we find the derivative of $\displaystyle \begin{align*} 3\arccot{(x)} + 3\arccot{ \left( \frac{1}{x} \right) } \end{align*}$?

To do this we should use implicit differentiation. If $\displaystyle \begin{align*} y = \arccot{(x)} \end{align*}$ then

$\displaystyle \begin{align*} \cot{(y)} &= x \\ \frac{\cos{(y)}}{\sin{(y)}} &= x \\ \frac{\mathrm{d}}{\mathrm{d}x} \left[ \frac{\cos{(y)}}{\sin{(y)}} \right] &= \frac{\mathrm{d}}{\mathrm{d}x} \left( x \right) \\ \frac{\mathrm{d}}{\mathrm{d}y} \left[ \frac{\cos{(y)}}{\sin{(y)}} \right] \, \frac{\mathrm{d}y}{\mathrm{d}x} &= 1 \\ \left\{ \frac{\sin{(y)}\left[ -\sin{(y)} \right] - \cos{(y)}\cos{(y)}}{\left[ \sin^2{(y)} \right] ^2 } \right\} \, \frac{\mathrm{d}y}{\mathrm{d}x} &= 1 \\ \left\{ \frac{ - \left[ \sin^2{(y)} + \cos^2{(y)} \right] }{\sin^2{(y)}} \right] \, \frac{\mathrm{d}y}{\mathrm{d}x} &= 1 \\ \left[ -\frac{1}{\sin^2{(y)}} \right] \, \frac{\mathrm{d}y}{\mathrm{d}x} &= 1 \\ -\csc^2{(y)} \, \frac{\mathrm{d}y}{\mathrm{d}x} &= 1 \\ - \left[ 1 + \cot^2{(y)} \right] \, \frac{\mathrm{d}y}{\mathrm{d}x} &= 1 \\ - \left( 1 + \left\{ \cot{ \left[ \arccot{(x)} \right] } \right\} ^2 \right) \, \frac{\mathrm{d}y}{\mathrm{d}x} &= 1 \\ - \left( 1 + x^2 \right) \, \frac{\mathrm{d}y}{\mathrm{d}x} &= 1 \\ \frac{\mathrm{d}y}{\mathrm{d}x} &= -\frac{1}{1 + x^2} \end{align*}$

Thus by the chain rule, we have

$\displaystyle \begin{align*} \frac{\mathrm{d}}{\mathrm{d}x} \left[ \arccot{ \left( \frac{1}{x} \right) } \right] &= \frac{\mathrm{d}}{\mathrm{d}x} \left[ \arccot{ \left( x^{-1} \right) } \right] \\ &= -x^{-2} \left[ -\frac{1}{1 + \left( x^{-1} \right) ^2 } \right] \\ &= \frac{ x^{-2} }{1 + x^{-2}} \\ &= \frac{1}{x^2 + 1} \end{align*}$

and thus

$\displaystyle \begin{align*} \frac{\mathrm{d}}{\mathrm{d}x} \left[ 3\arccot{(x)} + 3\arccot{ \left( \frac{1}{x} \right) } \right] &= -\frac{3}{1 + x^2} + \frac{3}{1 + x^2} \\ &= 0 \end{align*}$

 

[MarkFL]

We could also observe that:

\(\displaystyle \cot^{-1}(x)+\cot^{-1}\left(\frac{1}{x}\right)=\cot^{-1}(x)+\tan^{-1}(x)\)

This is a constant, \(\displaystyle \pm\frac{\pi}{2}\), for all real values of $x$, hence:

\(\displaystyle \frac{d}{dx}\left(\cot^{-1}(x)+\cot^{-1}\left(\frac{1}{x}\right)\right)=0\)