MHB Johnsy's question about finding a derivative via Facebook

  • Thread starter Thread starter Prove It
  • Start date Start date
  • Tags Tags
    Derivative
AI Thread Summary
The derivative of the function 3arccot(x) + 3arccot(1/x) is found to be zero. This is derived using implicit differentiation, where the derivative of arccot(x) is calculated as -1/(1 + x^2). The derivative of arccot(1/x) is also computed, resulting in 1/(x^2 + 1). Combining these results shows that the two derivatives cancel each other out, leading to a total derivative of zero. The relationship between arccot(x) and arctan(x) further confirms that their sum is a constant, thus reinforcing that the derivative is indeed zero.
Prove It
Gold Member
MHB
Messages
1,434
Reaction score
20
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

$\displaystyle \begin{align*} \frac{\mathrm{d}}{\mathrm{d}x} \left[ 3\arccot{(x)} + 3\arccot{ \left( \frac{1}{x} \right) } \right] &= -\frac{3}{1 + x^2} + \frac{3}{1 + x^2} \\ &= 0 \end{align*}$
 
Mathematics news on Phys.org
We could also observe that:

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

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

$$\frac{d}{dx}\left(\cot^{-1}(x)+\cot^{-1}\left(\frac{1}{x}\right)\right)=0$$
 
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...

Similar threads

Back
Top